Question

A 700-gram grinding wheel 22.0 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 215 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 50.0 s with constant angular acceleration due to friction at the axle.
What torque does friction exert while this wheel is slowing down?

Answer 1

Solution :

Given :

Mass of grinding wheel, m = 700 g

                                             = 0.7 kg

Diameter of the grinding wheel, d = 22 cm

                                                         = 0.22 m

Radius of the grinding wheel, r = 0.11 m

Initial angular velocity of grinding wheel, $\omega_0$ = 215 rpm

                                                                               $=215 \ rpm \times \frac{2 \pi \ rad}{1 \ rev}\times \frac{1 \ min}{60 \ s}$

where, $\pi = \frac{22}{7}$

Time taken to stop, t = 50 s

Final angular velocity is $\omega$ = 0

Angular acceleration of the grinding wheel is given by :

$\alpha = \frac{\omega-\omega_0}{t}$

   $=\frac{0-215 \ rpm \times \frac{2 \pi \ rad}{1 \ rev}\times \frac{1 \ min}{60 \ s}}{50 \ s}$

   $=-0.45 \ rad/s^2$

Magnitude of the angular acceleration of grinding wheel $\alpha$ $=-0.45 \ rad/s^2$

Moment of inertia of the grinding wheel (solid disk),

$I=\frac{1}{2}mR^2$

   $=\frac{1}{2} \times 0.7 \times 0.11^2$

  $=4.235 \times 10^{-3} \ kgm^2$

Torque exerted by friction while the wheel is slowing down is

$\tau = I \alpha$

  $=4.235 \times 10^{-3} \times 0.45$

 $=1.90 \times 10^{-3} \ Nm$