## Related questions with answers

A 50.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 160 N. The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of $6.50 \mathrm { N } \cdot \mathrm { m }$ between the axle of the stone and its bearings. How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s?

Solution

Verified### Concepts and Principles

1- **Rotational Acceleration** $\alpha$: The average rotational acceleration $\alpha$ (alpha) of a rotating rigid body (sometimes called angular acceleration) is its change in rotational velocity $\Delta \omega$ during a time interval $\Delta t$ divided by that time interval:

$\begin{gather*} \alpha = \dfrac{\Delta \omega}{\Delta t} \tag{1} \end{gather*}$

The unit for rotational acceleration is $\mathrm{(rad/s)/s}=rads/^2$.

2- **Kinetic Friction Force**: when an object slides along a surface, the components of the forces parallel to the surface that the surface exerts on the object are called kinetic friction forces. These forces oppose the motion of the object. The kinetic friction force depends on the surfaces themselves (on the coefficient of kinetic friction $\mu_k$) and relates to the magnitude of the normal force component $n$ of the total force that the surface exerts on the moving object as

$\begin{gather*} f_{k}=\mu_k n \tag{2} \end{gather*}$

3- **Rigid Object Under a Net Torque**: If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object undergoes an angular acceleration $\alpha$, where

$\begin{gather*} \sum \tau_{\mathrm{ext}} = I \alpha \tag{3} \end{gather*}$

This equation is the rotational analog to Newton's second law in the particle under a net force model.

4- The magnitude of the **torque** associated with a force $\overrightarrow{\mathbf{F}}$ acting on an object at a distance $r$ from the rotation axis is:

$\begin{gather*} \tau = r F \sin{\phi} = Fd \tag{4} \end{gather*}$

where $\phi$ is the angle between the position vector of the point of application of the force and the force vector, and $d$ is the moment arm of the force, which is the perpendicular distance from the rotation axis to the line of action of the force.

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