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A uniform cylinder of radius 10cmand mass 20kg is mounted so as to rotate freely about a horizontal axis that is parallel to and 5.0cm from the central longitudinal axis of the cylinder.

(a) What is the rotational inertia of the cylinder about the axis of rotation?

(b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

Short Answer

Expert verified
  1. The rotational inertia of the cylinder about the axis of the cylinder is 0.15 k²µ.³¾2.
  2. The angular speed of the cylinder as it passes through its lowest position is11 rad/s

Step by step solution

01

Given

The Radius of the cylinder is,R=10 c³¾=0.10″¾.

The mass of the cylinder is,m=20 k²µ.

The central longitudinal axis of the cylinder is, d=5.0 c³¾=0.05″¾.

02

Understanding the concept

Find therotational inertia of the cylinder about the axis of the cylinder using the parallel axis theorem. Then, using the law of conservation of energy, findtheangular speed of the cylinder as it passes through its lowest position.

Formula:

I=Icom+md2

Ei=Ef


03

(a) Calculate the rotational inertia of the cylinder about the axis of rotation

According to the parallel axis theorem,

I=Icom+md2

In this case,

I=12mR2+md2

Substitute all the value in the above equation.

I=12(20 k²µ)(0.1″¾)2+(20 k²µ)(0.05″¾)2I=0.15 k²µ.³¾2

Therefore, rotational inertia of the cylinder about the axis of the cylinder is,0.15 k²µ.³¾2 .

04

Step 4(b): Calculate the angular speed of the cylinder as it passes through its lowest position if the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates.

The cylinder is released from rest, let’s say, fromtheheight h.

So, its initial K.E is zero.

According to the conservation of energy,

Ei=Ef

K.Ei+P.Ei=K.Ef+P.Ef

In this case,

0+mgh=12±õÓ¬2+0Ó¬2=2mghI

Substitute all the value in the above equation.

Ó¬=2(20 k²µ)(9.8″¾/s2)(0.050″¾)(0.15 k²µ.³¾2)Ó¬=11.4 r²¹»å/s≈11 r²¹»å/s

Therefore, angular speed of the cylinder as it passes through its lowest position is11 r²¹»å/s .

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