Question

the moment of inertia of grindstone about its axis of rotation is 25kgm².Starting from rest it acquires a speed of 120rpm in 10 secs. Find the torque acting on it

Answer 1

Answer:5Nm

Explanation:torque = IdL/dt i.e rate of change in angular momentam

L initial =0

L final =120revolutions per minute

Convert to revolutions per sec

120/60 =2rps

Applying above formula

Torque = 25*(2—0)/10sec

Ans 5Nm

Answer 2

Torque, $\tau=31.4\ N-m$

Explanation:

Given that,

Moment of inertia of grindstone about its axis of rotation is, $I=25\ kg-m^2$

Initial speed of the grindstone, u = 0

Final speed of the grindstone, v = 120 rpm = 12.56 rad/s

Time, t = 10 s

We need to find the torque acting on it. The relation between the torque and moment of inertia is given by :

$\tau=I\alpha$

$\tau=I\dfrac{v-u}{t}$

$\tau=25\times \dfrac{12.56-0}{10}$

$\tau=31.4\ N-m$

So, the torque acting on it is 31.4 N-m. Hence, this is the required solution.

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