Question

A metal ring of mass 5 g and
radius 1 cm can rotate in
horizontal plane about
tangent normal to plane of
ring. The ring is initially at
rest. Due to constant torque,
its angular velocity becomes
10 rad/sec in 10 seconds.
The value of torque is------​

Answer 1

Answer:

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Explanation:

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Answer 2

Given:

Mass of the given  metal ring = 5 g

Radius of  the given metal ring = 1 cm

Time taken for the angular velocity of the metal ring to become 10 rad/sec starting from rest = 10 sec

To Find :

The value of torque applied = ?

Solution :

We can find the angular acceleration of the given metal ring by using the given values from the  equation :

$\omega = \omega_0 + \alpha t$

Since initially the ring was at rest , so $\omega_0$  = 0

So, $10 = 0 + 10\alpha$

Or, $\alpah = \frac{10}{10}$

So,$\alpha = 1 \frac{rad}{sec^2}$

Now the moment of inertia of the given metal ring about its centre is :

$I = MR^2$

 = $5 \times 10^{-3} \times (10^{-2})^2$

 = $5 \times 10^{-7}$ kg-m

Now the torque on the given metal ring  can be given as :

$\tau = I\alpha$

  =$5 \times 10^{-7} \times 1$

  =$5 \times 10^{-7}$ kg-m

Hence the value of torque applied on the metal ring is $5 \times 10^{-7}$ N-m .