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Rotation Lab Quiz
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Gravity
Terms in this set (30)
One complete revolution corresponds to 360 degrees which corresponds to
2pi radians
For an object rotating about an axis, how is the frequency (f) of the motion related to the angular velocity (w)?
w = 2
pi
f
This term describes the action of a rigid body as it rotates about some fixed axis.
rotational motion
The rate of change of angular velocity is known as ______ and has units of ________.
angular acceleration; rad/s^2
The tendency of an object to oppose a change in its state of relative rest or constant velocity and measured by its mass is called
inertia
Assume the following objects have the same mass which has the greatest moment of inertia?
hoop
According to Newton's 2nd Law the rotational analogue of force is
torque
The product of moment of inertia times angular acceleration is called
torque
Given a continuous distribution of mass and a recognizable geometry the definition of I (moment of inertia) is the integral of the product of elementary masses by the square of respective distances from a selected axis of rotation.
true
One defintion of torque is the product of __________ times ________.
Force; perpendicular distance to axis of rotation
Suppose you apply a torque of equal magnitude to two disks which have the same moment of inertia I but in which the radius of the first disk (r1) is less than the radius of the second disk (r2). Which of the following statements is true?
alpha1 (angular acceleration of disk 1) = alpha2 (angular acceleration of disk 2)
Suppose you apply a torque of equal magnitude to two discs which have the same moment of inertia I but in which the radius of the first disc (r1) is less than the radius of the second disc (r2). Which of the following is true about the tangential acceleration on the edge of the disk?
a1 is less than a2
For a disc of radius 10.0 cm used on the Rotational Dynamics Apparatus a frequency of 50.0 Hz was recorded. For the one-second time interval on the RDA what distance (in cm) passed in front of the counter?
3140 cm
For a disc of radius 10.0 cm used on the Rotational Dynamics Apparatus a frequency of 50.0 Hz was recorded. For the one-second time interval on the RDA what was the angular displacement of the disc in radians?
314
For a disc of radius 10.0 cm used on the Rotational Dynamics Apparatus a frequency of 50.0 Hz was recorded. For the one-second time interval on the RDA what was the angular displacement of the disc in degrees?
1.80E04 degrees
For a disc of radius 10.0 cm used on the Rotational Dynamics Apparatus a frequency of 50.0 Hz was recorded. For the one-second time interval on the RDA what was the average angular velocity (rad/s) of the disc during that second?
314 rad/s
For a disc of radius 10.0 cm used on the Rotational Dynamics Apparatus a frequency of 50.0 Hz was recorded. If the mass of the disc is 200.0 grams what is the moment of inertia of the disc (gram*cm^2)?
1.00E04 g*cm^2
In m/s what is the linear speed of a point at the equator on the surface of the earth? Assume the radius of the earth to be 6400 kilometers. HINT: the earth rotates 2*Pi radians in 24.0 hours OR one rotation per day OR 360. degrees in 24.0 hours.
470 m/s
A steel disc with moment of inertia 7.8 kg*m^2 has an angular acceleration of 3.10 rad/s^2. What is the torque exerted on the disc?
24 N*m
A string wrapped around the circumference of a disc has a hanging mass attached to it. The tension in the string is 3.9 N. The diameter of the disc is 2.5 m. What is the torque applied to the disc?
4.9 N*m
For a steel disc of mass M equal to 1.30 kg and radius R equal to 0.660 m, a torque of 5 N*m is applied. What is the angular acceleration of the disc?
17.7 rad/s^2
What is the slope of the velocity of hanger vs. time graph?
The acceleration
Why did you have to experimentally find the moment of inertia for the ring and disk together?
It is used to find the moment of inertia for the ring by itself.
When finding the friction mass, the system should rotate at a constant angular velocity...
after a small initial push
True or False. The moment of inertia of disk rotated about its center is less than the moment of inertia of the disk rotated about its diameter.
false
The torque applied to the disk is equal to:
Moment of inertia of the disc times the angular acceleration of the disc
Why was both the inner and outer radius of the ring taken?
Needed it to find its moment of inertia using I=.5M(R(inner)^2+R(outer)^2)
If the radius of the pulley were increased, the torque applied to the disc would:
increase
If the moment of inertia of the disc were increased and the torque remained constant, the angular acceleration of the disc would:
decrease
If the moment of inertia of the disc were increased, the torque applied by the hanging mass would:
remain the same
;