A wheel with radius 0.33 m and rotational inertia 2.0 kg⋅m2 spins on an axle with an initial angular speed of 3.0 rad/s. Friction in the axle exerts a torque on the wheel, causing the wheel to stop after 6.0 s. The average torque exerted on the wheel as it slows down has magnitude
A) 0.50 N⋅m
B) 1.0 N⋅m
C) 2.0 N⋅m
D) 3.0 N⋅m

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shubhamchouhanvrVT shubhamchouhanvrVT · Answer 1 · 2022-03-29T17:53:32+02:00

The average torque exerted on the wheel will be [tex]T=1\ N-m[/tex]

Angular torque will be defined as the product of the moment of inertia of the rotating body and the angular acceleration of the body.

Now it is given in the question that

Angular velocity [tex]w=3\ \frac{rad}{s}[/tex]

Time t=6 sec

Moment of inertia [tex]I=2\ \frac{kg}{m^2}[/tex]

The angular torque will be given by the formula

[tex]T=I\alpha[/tex]

here the angular acceleration will be given as

[tex]\alpha =\dfrac{\Delta w}{t}[/tex]

[tex]T=\dfrac{I\Delta w}{t}[/tex]

[tex]T=\dfrac{2\times3}{6} =1\ N-m[/tex]

Thus the average torque exerted on the wheel will be [tex]T=1\ N-m[/tex]

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