94. The moment of inertia of the empty turntable is 1.5kgm2. With a constant torque of 2.5Nm, the turntable–person system takes 3.0sto spin from rest to an angular speed of 1.0rad/s. What is the person’s moment of inertia about an axis through her center of mass? Ignore friction in the turntable axlerole="math" localid="1667995844298" (a)2.5kgm2;(b)6.0kgm2;(c)7.5kgm2;(d)9.0kgm2.

Short Answer

Expert verified

The angular momentum of the person is Iperson=6.0kgm2. Hence, option b is correct.

Step by step solution

01

Given Data

It is given that the moment of inertia of empty turntable as I0=1.5kgm2, constant torque as τ=2.5Nm, time as t=3.0sand final angular speed as ω=1.0rad/s.

02

Angular acceleration

The angular acceleration is produced by the constant torque as τ=Iα…… (1)

Find afrom the equation ω=ω0+αt. Since, the initial angular speed is zero that is ω0=0thenα=ωτ.

Substitute αin (1) then, τ=Iωτ.

03

Angular momentum of the system

The total angular momentum of the system is I=I0+Iperson. Then τcan be written as

τ=I0+IpersonωtI0+Iperson=rtωIperson=rtω=I0

Substitute all the known values and simplify.

Iperson=τtω-I0=2.5Nm3.0s1.0rad/s-1.5kgm2=7.5kgm2-1.5kgm2=6.0kgm2

Therefore, the angular momentum of the person is Iperson=6.0kgm2.

Hence, option b is correct.

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