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

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?

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Thu Giang 5 years 2021-08-20T04:19:43+00:00 1 Answers 17 views 0

Answers ( )

    0
    2021-08-20T04:21:35+00:00

    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

    $\tau = I \alpha$

      $=4.235 \times 10^{-3} \times 0.45$

     $=1.90 \times 10^{-3} \ Nm$

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