Question

A 800-gram grinding wheel 27.0 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 245 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?

Answer 1

Explanation:

d = Diameter of wheel = 27 cm

r = Radius = $\frac{d}{2}=\frac{27}{2}=13.5\ cm$

m = Mass of wheel = 800 g

$\omega_i$ = Initial angular velocity = $245\times \frac{2\pi}{60}\ rad/s$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-245\times \frac{2\pi}{60}}{50}\\\Rightarrow \alpha=-0.51312\ rad/s^2$

Moment of inertia is given by

$M=\frac{1}{2}mr^2\\\Rightarrow M=\frac{1}{2}\times 0.8\times 0.135^2\\\Rightarrow M=0.00729\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.00729\times -0.51312\\\Rightarrow \tau=-0.0037406448\ Nm$

The torque the friction exerts is -0.0037406448 Nm

For more information on torque and moment of inertia refer

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