Question: A disk with a rotational inertia of 7.00 kg m2rotates like a merry-go-round while undergoing a variable torque givenbyτ=(5.00+2.00t)Nm. At time, t = 1.00s its angular momentum is 5.00 kg m2 /swhat its angular momentum is at t =3.00s

Short Answer

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Answer

The angular momentum of the disk at t = 3.00 isL=23kgm2/s

Step by step solution

01

Given

  1. The moment of inertia of disk,I=7.00kgm2.
  2. The torque of the rotating disk,T=5.00+2.00tNm.
02

To understand the concept

Using the relation between torque and angular momentum, we can find the angular momentum from torque.

Formula:

τ=dLdt

03

Calculate the angular momentum as a function of time

We know that,

T=dLdt

Therefore, the angular momentum as a function of time is

Lt=TdtLt=(5+2t)dtLt=5t+2t22+cLt=5t+t2+c

(Where c is constant of integration)

We are given thatL=5att=1s.

Lt=5t+t2+c5=51+12+c5=6+cc=-1

Therefore, angular momentum is

Lt=5t+t2-1

04

Calculate the angular momentum at t =3.00 s

Therefore, the angular momentum of the disk at t =3.0 s is

Lt=3=53+32-1L=15+9-1L=23kgm2/s

Angular momentum of the disk at t =3.00s is23kgm2/s

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Most popular questions from this chapter

Figure is an overhead view of a thin uniform rod of length 0.600mand mass Mrotating horizontally at 80.0rad/scounter clock-wise about an axis through its centre. A particle of mass M/3.00and travelling horizontally at speed 40.0m/shits the rod and sticks. The particle’s path is perpendicular to the rod at the instant of the hit, at a distance dfrom the rod’s centre.

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