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07_ Torque
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Gravity
Key Concepts:
Terms in this set (56)
torque
rotational force; any action that causes an object to rotationally accelerate[rotate faster|| increase in rotational speed];
any action that causes a change to the rotational speed of an object.
as force causes acceleration, torque cause rotational acceleration.
as force causes an object to accelerate; torque causes and object to rotate;
what are the torque parameters?
(1) torque depends on two things. the force applied and the distance from the axis of rotation[lever arm] to the force.
(1) depends on the magnitude of the force and the distance between the force & [axis] point of rotation.
(2) only the perpendicular component of a force contributes to the torque. Now this is true but sometimes we can calculate torque in a different way. Specifically using the lever arm, because it doesn't matter where we apply the angle.
(3) counterclockwise rotation corresponds to + torque & clockwise rotation corresponds to - torque.
(4) if net torque is zero an object is either not rotating or rotating at a constant speed. if net torque doesn't equal zero the object experiences a change in its rotational speed.
(5) the origin, is usually chosen to coincide with the axis of rotation but any point can be chosen
Rotational analog of Newton's 1st Law
the rate of rotation of an object doesn't change unless the object is acted on by a net torque.
when solving for net torque what determines whether you add or subtract an individual value of torque within the net torque?
the right-hand rule
or
counterclockwise torques are added and clockwise torques are subtracted.
unit of torque
the SI unit of torque is the Newton meter (not to be confused with the Joule)
right-hand rule
(1) Point the fingers of your right hand in the direction of position vector r
(2) Curl your fingers toward the force vector F.
(if you are able to curl your fingers natural towards vector F, your thumb points in the direction of torque, which should he out of the plane of the page which corresponds to positive value of torque or counterclockwise rotation)
(if you are unable to naturally curl your fingers towards vector F, flip your hand 180 degrees and point in direction of position vector r. You should now be able to curl your fingers natural. negative torque, clockwise rotation)
If the applied force is not perpendicular does torque still exists?
Yes torque still exists with the exception of applied forces parallel to the position vector.
the trig function sin theta is now used to isolate the component of the torque that is perpendicular to position vector.
lever arm
[1]the perpendicular distance from the axis of rotation to a line of action[line where the force acts]
[2] perpendicular distance from the axis of rotation to an imaginary line along which the force acts[force line is a good synonmy for line of action]
Under what two conditions will lever arm be used?
(1) when the Force acting on the object is not perpendicular to its' position vector
(2) when you have multiple points not all along the same straight line (position vector) and no consistent line where all force act. in this scenario lever arm would allow you to solve for multiple torques as if they were all acting upon the same line.
True or False
Torques values must be calculated from the specific axis of rotation.
False. torques can be calculated from any point. In fact torques can even be calculated even if the object is not rotating at all.
what two conditions must an object satisfy to be in mechanical equilibrium?
when are these conditions true?
the net external force and net external torque must equal zero.
these conditions are true if the object is moving or rotating at constant speed, or when the object is not moving or rotating at all.
REMINDERS: Solving Torque
(1) Because any point can be chosen as the origin for the position vector, it is best to choose a point that has a force already acting so we can eliminate that force from net torque equation
(2) for an object sitting on top of another at one point there will be a contact(normal) force at that point. [Ex: fulcrum point] if the two objects are in contact at many points we can say that the net contact(normal) force would be located at a symmetrical point between the two objects.
(3) remember that weight and mg are different forces that act on an object. Weight is actually normal force upwards and mg is force due to gravity.
[IMAGE]
Consider an object of arbitrary shape lying in the x-y plane allowed to rotate about an origin. If the object is divided into a large number of very small particles (each contributing a torque equal to its weight times its lever arm) what would this be equivalent to?
[ANSWER]
Consider an object of arbitrary shape lying in the x-y plane allowed to rotate about an origin. If the object is divided into a large number of very small particles (each contributing a torque equal to its weight times its lever arm) what would this be equivalent to?
The effect of rotation on the object of the individual particles would be equivalent to a single force of magnitude mg applied at the singular point called the object's center of gravity.
What's the difference between center of mass and center of gravity?
If the body(object) is so large that gravity varies significantly over the body, then the center of gravity will be different than the center of mass.
Except for this scenario, there is no appreciable difference between center of gravity and center of mass.
Under what conditions will the center of mass lie on the axis of symmetry?
Bodies the are homogeneous and symmetrical.
Equation for Center of Mass
What condition(s) must be true for an object to be in equilibrium?
Net force in x & y direction must equal zero and net torque must also equal zero
PROBLEM SOLVING STRATEGY
Objects in Equilibrium
[IMAGE]
How do we know that there is a downwards force R acting at the joint(point O)?
what force is missing?
[ANSWER]
How do we know that there is a downwards force R acting at the joint(point O)?
what force is missing?
Technically, the force R acting at the joint is the force due to gravity on the humerus [mg(weight) of humerus]. Not pictured is the force due to gravity of the ulna(mg of ulna) but this specific problem/image neglects to include this force though it too is also acting.
[IMAGE]
How can point R [pivot point on a wall] have an x & y component?
[ANSWER]
How can point R [pivot point on a wall] have an x & y component?
Firstly, lets establish that there is a pin in the beam at the wall, when considering rotational motion, the beam can rotate at two different locations. If the rope was not attached to the right end of the beam, the beam could rotate around the pin(left end). The pin would be the origin of this rotation. If the left end of the beam has its pin removed, the beam would rotate around the right end (rope-beam attachment point) . The right end (rope-beam attachment point) would be the origin of rotation. The point R has an x-component because the pin exerts a force leftward to counteract the rope rightward tension force pushing the beam to the right. The pin exerts an upward force R to counteract the force that wants the left side of the beam to fall down( or rotate clockwise)
REVISITING PAST CONCEPTS
how should we think to determine the direction of static friction?
For an object not moving one should think if there were no static friction, which direction would the object move. Then you should know that the Force of static friction acts in the direction opposite
For an object moving as the result of static friction, think the static friction points in the direction due to inertia (I think)
[IMAGE]
Proof of the relationship between torque and angular acceleration. Let's assume that a Force (Ft) was applied to an object of mass m perpendicular(or tangentially) to the rod distance r from the origin(axis of rotation)
[ANSWER]
Proof of the relationship between torque and angular acceleration. Let's assume that a Force (Ft) was applied to an object of mass m perpendicular(or tangentially) to the rod distance r from the origin(axis of rotation)
moment of inertia
(1)the mass property of a rigid body that determines the torque needed for a desired angular rotation. (2) moment of inertia is the sum of each mass particle times the square distance to the axis of rotation.
rotational analog of Newton's 2nd law
angular acceleration of an extended rigid object is directly proportional to net torque and inversely proportional to the moment of inertia
When comparing a smaller gear to a larger gear:
(a) does a smaller gear increase or decrease torque? why?
(b) does a smaller gear increase or decrease bike speed? why?
(a) Torque decrease because the radius is now smaller. Torque is defined as radius times Force. If the applied force is the same but you decrease the radius, the value of Torque also decreases.
(b) Bike speed increases. Lets consider two bikes. Bike A has two gears of equal size. Bike B has one gear that is the same size as bike A and its other gear[the rear gear] is half its size. Lets assume that each bike's gears are connected to each other via chain. Lets also assume that the second gear is fastened to back wheel so that the gear and wheel rotate the same number of times. Said another way if the back gear rotates three times the back wheel also rotates three times. In bike A, for every one rotation of the first gear the second gear & wheel also rotates once. In bike B for every one rotation of the first gear the smaller gear & WHEEL rotates twice. Since the smaller gear is fastened to the axle, the gear and wheel rotate the same number of times. If the time it takes for the rotation of the first gear in both bikes is the same the 2nd bike will travel further during that time period because the wheel[gear] rotates twice. Since speed is the distance traveled over time and the time is constant the 2nd bike moves faster.
Explain how force, torque, rotational speed, and angular acceleration relate to a 5-speed bike. Assume that the bike starts from rest.
A constant force is applied to a pedal causing it to rotate. The pedal is fastened to a gear so that the gear and pedal rotate at the same speed. The pedal gear is connected to a rear gear via a chain. In the same way that the first gear is fastened to the pedal so to is the rear gear attached to the back wheel[via the axle]. As a result, the rear gear and rear wheel will also rotate at the same speed. Since the bike was initially at rest the gears[& wheels] all experience an increase in rotational speed until a maximum value is reached that depends on the tangential chain speed. Once this maximum chain speed is reached, the bike will not move any faster on this gear. The chain is then moved to the next gear witch has a smaller radius. This action has two effects. (1) The tangential speed of the chain decreases & (2) the bike becomes more difficult to pedal. The chain speed decreases because of the decrease in gear radius. Tangential speed is the product of the radius & (rotational) angular speed. Both gears have the same rotational speed because they are fixed into the axis. However the second gear is smaller and thus has a smaller radius value, thus a smaller tangential speed. As for the 2nd effect, the bike becomes more difficult to pedal because more Force is required to get the object to rotate. Torque equals the product of Force & radius. A bike requires a certain torque value to achieve the maximum tangential chain speed. Let's consider a hypothetical bike where 10 units of torque are needed to get to the maximum tangential speed. When we shift to the smaller gear the value of torque decreases. [For our hypothetical bike, let's say it decreases to 6 units of torque]. If the rider wants to maintain the 10 units of torque that he was at before shifting gears he must now exert a larger force because he is limited by the fact that the radius of the new gear is smaller. If he were to continue to exert the same force, the bike would slow down. To get back up to the 10 units of torque, the rider exert more force [or experiences more resistance] on the pedal.
[IMAGE]
Is the FBD drawn correctly for the pulley in part (a)?
[ANSWER]
Is the FBD drawn correctly for the pulley in part (a)?
No it is not. The FBD are drawn with respect to the each object. If we were to draw the FBD for the pulley, the tension forces would point in the opposite direction as currently represented.
Identify pseudonyms for the following terms:
(a.) angular velocity
(b.) angular acceleration
(a) rotational velocity or velocity of rotation
(b) rotational acceleration or acceleration of rotation.
Consider the following passage:
"When a diver or an acrobat wishes to make several somersaults, she pulls her hands and feet close to the trunk of her body in order to rotate at a greater angular speed. IN THIS CASE, THE EXTERNAL FORCE DUE TO GRAVITY ACTS THROUGH HER CENTER OF GRAVITY AND HENCE EXERTS NO TORQUE ABOUT HER AXIS OF ROTATION, SO THE ANGULAR MOMENTUM ABOUT HER CENTER OF GRAVITY IS CONSERVED. For Example, when a diver wishes to double her angular speed, she must reduce her moment of inertia to half its initial value."
Explained the capitalized statement. How does the diver differ from the figure skater? Explain how no torque is acting on the skater and net torque is zero for the diver because gravity acts through her center of gravity.
No torque is exerted because the force due to gravity on the diver is acting at the origin and thus has a radius of zero. So the force due to gravity to at 0 distance from axis of rotation is zero torque... The diver differs from the skater in that the diver does experience a net force that could affect rotation, gravitational force. However, this force does not affect rotation because the force acts at the center of rotation which means net torque for the system is zero. The skater is already spinning and has no torque acting on her to increase or decrease her rotation. So net torque is zero and the law of conservation of angular momentum still applies. She did exert a force initially to get her to spin, but that force is no longer acting at the point where she varies her arms to change her rotational speed.
QQ 8.1 Using a screwdriver, you try to remove a screw from a piece of furniture but can't get it to turn. To increase the chances of success you should use a screwdriver that (a) is longer (b) is shorter (c) has a narrower handle or (d) has a wider handle.
Correct answer is (d).
A wider handle is equivalent to increasing your radius (distance from the axis of rotation). Since torque is defined as radius times Force, the same Force with a wider handle generates a large torque, or larger rotational Force. .The more torque you have the easier it is to turn.
Consider opening a revolving door. Where on the door would it be easiest to apply a force to open the door?
At the door's hinge or at the door's knob.
Explain why this is so.
At the knob. The knob is further away from the axis of rotation and therefore a given force will generate the largest torque (rotating force)
For a given Torque, what happens when the radius or force is changed? What happens to torque if the Force is constant but the radius changes?
For a given torque(rotational force) if you increase the radius(distance from the axis of rotation) less force is needed to produce the torque.
If you increase the radius and keep the Force the same, the amount of torque increases.
[IMAGE]
QQ8.3 The two rigid objects shown in Figure 8.16 have the same mass, radius and angular speed. If the same braking torque is applied to each, which takes longer to stop (a) A (b) B (c) more information is needed.
[ANSWERED]
QQ8.3 The two rigid objects shown in Figure 8.16 have the same mass, radius and angular speed. If the same braking torque is applied to each, which takes longer to stop (a) A (b) B (c) more information is needed.
the correct answer is (B). while both objects have the same mass, the mass of the ring is concentrated further away from the axis of rotation and thus has a larger torque. So a larger breaking torque will be required to stop that object.
See analogous [IMAGE]
Rotational Kinetic Energy
proof and equation
How does rotational kinetic energy change the Work-Energy theorem?
What are the new EQUATIONS
Translational & Rotational kinetic energy are now incorporated. Generally speaking, Some kinetic energy is lost as rotational KE.
PROBLEM SOLVING STRATEGY
Energy & Rotation
angular momentum
momentum of rotation; moment of inertia times angular velocity
conservation of angular momentum
How is an isolated system now defined?
an isolated system is now defined as the state where mechanical energy, linear momentum and angular momentum remain constant
Explain conservation of momentum through the example of a spinning figure skater
When the skater pulls her arms and legs close to her body, this reduces their distance from her axis of rotation and hence also reducing her moment of inertia. Because angular momentum is conserved, a reduction in her moment of inertia must increase the angular velocity. Coming out of the spin she needs to reduce her angular velocity so she extends her arms and legs again thereby increasing her moment of inertia and slowing her rotation.
(1)How does a massive star collapsing under the influence of gravitational force demonstrate concepts of angular momentum?
(2)How does this cause a gigantic outburst of energy supernova)?
Any star whether it is a planet, sun, or moon rotates about its own axis. Just like the earth rotates about itself so to do stars, rotate about their own axis. A star like the sun has energy. This energy is provided by the nuclear fusion reaction of helium and hydrogen. When a star runs out of energy, gravitational forces cause the star to collapse upon itself. Since momentum(specifically angular momentum) is conserved, the star rotates at a faster rate because the mass has moved closer to the origin.
when the star collapses upon itself all the particles come into contact and ignite a large nuclear fusion reaction. this fusion reaction results in a large gigantic outburst of energy(supernova).
Two interesting facts about center of gravity
(1) an object will be balanced(meaning the object will not rotate or tip over) if it is supported at the center of gravity or any point on the vertical line above or below it;
(2) if an irregularly shaped object is hung or supported from any position (or point), the objects center of gravity will always lie somewhere on the straight line above or below the position or point of support.
physics behind the high jump and the Fosbury flop
pre-Fosbury, jumpers had to get their entire center of mass over the bar in order to clear it.
With the Fosbury flop technique, the jumper gets his entire body over the bar while his center of mass moves under it.
To explore this idea further lets break the jumper into three main parts( head, torso/back, & legs. At the initial instance of the jump, the jumpers head is over the bar while his back and legs are not. At the middle instance, his back is over the bar while his head and legs are not. At the last instance, his legs are over the bar but his back and head are not. If you were to average where his center of mass was, it would be under the bar.
Since the center of mass can be lower but still allows the jumper to clear the bar, less vertical force is necessary for the jump. In other words an equivalent non-Fosbury vertical force allows a Fosbury jumper to clear a higher bar than the non-Fosbury jumper.
[DIAGRAM]
How does curving your body move your center of gravity outside your body?
[EXPLANATION]
How does curving your body move your center of mass outside your body?
The position of a persons center of mass changes all the time with movement. A standing person has a center of mass inside the body 2/3 up from the ground. If the individual were to lift there hands, the center of mass moves higher because some mass is now redistributed at a higher position. when a person bends over and touches their toes, the center of mass shifts outside the body. Like V shaped objects.
QQ8.5 Which arrives at the bottom first? (a) a ball rolling without sliding down an incline (b) a solid cylinder rolling without sliding down the incline, (c) a box of the same mass as the ball sliding down a frictionless incline having the same dimensions as A? Assume each object is released from rest at the top of its incline.
(c) In option C, the initial mgh is converted to translational KE, while in option A & B, some of the initial mgh is converted into rotational KE and lessens the value of translational at the end.
The ball would be second to reach the end because its moment of inertia is less than a solid cylinder resulting in a smaller rotational KE at then end. As a result the translational KE would be slightly larger for the same initial PE.
QQ8.7 If global warming continues, it's likely that some ice from the polar ice caps of the Earth will melt and the water will be distributed closer to the Equator. If this occurs, would the length of the day (one revolution) (a) increase, (b) decrease, or (c) remain the same?
(a) Increase.. Since angular momentum is conserved, the rotational speed would decrease due to the redistribution of mass from poles to equator. There is a inversely proportional relationship between rotational speed and Time for a given arc length traveled according. Therefore, the period increase with decreasing rotational speed.
CQ1 Why can't you put your heels ﬁrmly against a wall and then bend over without falling?
In order for you to not fall, your center of gravity must be on the same line as your support point, your feet. If your heals are against the wall, when you bend over your center of gravity is no longer on the same line(directly above) your support point. As a result you lose your balance.
CQ7 In some motorcycle races, the riders drive over small hills, and the motorcycle becomes airborne for a short time. If the motorcycle racer keeps the throttle open while leaving the hill and going into the air, the motorcycle's nose tends to rise upwards. Why does this happen
As the motorcycle leaves the ground, the friction between the tire and the ground suddenly disappears. if the motorcycle driver keeps the throttle open while leaving the ground, the rear tire will increase its angular speed and, hence its angular momentum. The airborne motorcycle is now an isolated system, and its angular momentum must be conserved. The increase in angular momentum of the tire directed, say clockwise must be compensated for by an increase in angular momentum of the entire motorcycle counterclockwise. This rotation results in the nose of the motorcycle rising and the tail dropping.
The higher the gear the faster a bike can move. Why is it difficult to start pedaling a bike in a higher gear?
A bike in a higher gear requires more Force because its radius is smaller. Lets consider two bikes in different gears. Let Bike A be in first gear and bike B in third. Gears are fixed to the axle of a wheel so that the gear and wheel rotate at the same rate. Torque causes the gear (and wheel) to rotate. Lets say that 10 units of torque are need to get the gear to spin its fastest. Torque equals force times radius. The larger gear has a bigger radius so less force is needed to achieve the 10 units of torque. The small gear has a smaller radius so more force is required to get to the 10 units of torque.
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