Question

6. Speed of a body spinning about an axis increases from rest to 100 rpm in 5 second if a constant torque is applied. The external torque is then removed and the body comes to rest in 100sec due to friction Find the frictional torque. ​

Answer 1

The frictional torque is 1.002Nm.

Given:

an axis increases from rest to 100 rpm in 5 second

The external torque is then removed and the body comes to rest in 100sec

To find:

The frictional torque.

Solution:

Here,

Initial frequency, f*1 = 0 rpm (starting from rest)

After t1 = 5 sec, frequency, f2 = 100 rpm = (100/60) rps

Torque, T = 20 Nm

After time interval, t2 = 100 sec

Final frequency, f3 = 0 rpm = 0 rps

Frictional torque, tau_{F} =?

We know that,

(a) Angular acceleration, alpha = w2- w1/t

Or, alpha = (2π*f_{2} - 2π*f_{1})/t

Or, alpha = (2π*f_{2} + 0)/t

Or, alpha = (2π)/5 * 100/60

Or, alpha = 2.094 rad / (s ^ 2)

Now, Torque, t = I*alpha

Or,I= tau /'alpha = 20/2.094 = 9.55kg * m ^ 2

Again, when torque is removed the body comes to rest in 100 sec

Then angular retardation due to friction

alpha_{t} = (w3 - w2)/t2

Or alpha_{t} = (2π*f3- 2π*f2)/t2

Or alpha_{t} = (0 - 2pi*f_{2})/t2

Or, alpha_{f} = - (2pi)/100 * 100/60

Or, alpha r = - 0.105 rad sec^ -2

.. Frictional torque, Tf, = 9.55 x 0.105 = 1.002 Nm

Therefore, The frictional torque is 1.002Nm.

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