Question

The moment of inertia of a disc of radius 0.3m about its geometric axis is 2Kg-m2. If a string is tied to its circumference and a force of 15 Newton is applied,
value of angular acceleration with respect to this axis will be :-
​

Answer 1

Answer:

your answer is 5N-m

Explanation:

Correct option is

B

5 N-m

Torqueabouttheaxisisτ=r×F

inthiscaser=0.5andF=10N

ThereforeTorque=0.5×10

=5N−m

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Answer 2

Given: The moment of inertia of a disc of radius 0.3m about its geometric axis is 2Kg-m2 and force applied is 15 N.

To find: Value of angular acceleration.

Solution:

• Torque or angular acceleration is the force that tends to cause rotation.
• Torque about the axis is given by,

        $\tau = r * f$

• Here, τ is the torque or the angular acceleration of the string, r is the radius of the disc and f is the force applied to the string.
• Hence, the angular acceleration or the torque is,

        $\tau = 0.3m * 15N$

            $=4.5 Nm$

• Moment of inertia is mass times the square of the perpendicular distance to the axis of the body.
• Although given in the question, it is not required for the calculation of angular acceleration.

Therefore, the value of angular acceleration with respect to this axis is 4.5 Nm.