A700-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?

0.01524 Nm

Explanation:

= Final angular velocity = 0

= Initial angular velocity = 210 rpm

= Angular acceleration

= Angle of rotation

t = Time taken = 41 seconds

Converting rpm to rad/s

m = Mass of wheel = 700 g

r = Radius of wheel = 28.5 cm

Equation of rotational motion

Moment of inertia of a disc is given by

Torque is given by

The torque the friction exerts while slowing down is 0.01524 Nm

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.