Question

A 700-gram grinding wheel 28.5 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 210 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 41.0 s with constant angular acceleration due to friction at the axle. Part A What torque does friction exert while this wheel is slowing down?

Answer 1

Answer:

0.01524 Nm

Explanation:

$\omega_f$ = Final angular velocity = 0

$\omega_i$ = Initial angular velocity = 210 rpm

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken = 41 seconds

Converting rpm to rad/s

$210\times \frac{2\pi}{60}$

m = Mass of wheel = 700 g

r = Radius of wheel = 28.5 cm

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-210\times \frac{2\pi}{60}}{41}\\\Rightarrow \alpha=-0.53636\ rad/s^2$

Moment of inertia of a disc is given by

$I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}\times 0.7\times 0.285^2\\\Rightarrow I=0.02842875\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.02842875\times 0.53636\\\Rightarrow \tau=0.01524\ Nm$

The torque the friction exerts while slowing down is 0.01524 Nm