Question

A constant net torque equal to 20 N-m is exerted on a pivoted wheel for 8 sec, during which time the angular velocity of the wheel increases from zero to 100 re//min. The external torque is then removed and the wheel is brought to rest by friction in its bearings in 70 sec. Compute (a) the moment of inertia of the wheel about the rotation axis, (b) the friction torque(c)thetotal no. of revolutions made by the wheel in the70 sec time interval

Answer 1

Answer:

Part a)

Moment of inertia of the wheel is

$I = 15.3 kg m^2$

Part b)

Torque due to friction force is given as

$\tau_f = -2.29 Nm$

Part c)

Number of revolutions are given as

$N = 58.5 rev$

Explanation:

Part a)

As we know that when torque is applied on the wheel then its angular speed increases from its rest position

So here we will have

$\tau \times t = I (\omega_f - \omega_i)$

now we know that

$\tau = 20 Nm$

t = 8 s

Final frequency of revolution is 100 rev/min so we have

$\omega_f = 2\pi(\frac{100}{60})$

$\omega_f = 10.5 rad/s$

now we know by above equation

$20 (8) = I (10.5 - 0)$

$I = 15.3 kg m^2$

Part b)

As we know that due to frictional torque it comes to rest in 70 s

so we have

$\tau_f = \frac{I(\omega_f - \omega_i)}{\Delta t}$

$\tau_f = \frac{15.3(0 - 10.5)}{70}$

$\tau_f = -2.29 Nm$

Part c)

Now by kinematics we can find the number of revolutions

so we have

$N = \frac{\omega_f + \omega_i}{4\pi} t$

$N = \frac{10.5 + 0}{4\pi} 70$

$N = 58.5 rev$

#Learn

Topic : Torque and inertia

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