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Question: A uniform solid cylinder of mass \[5\,{\text{kg}}\] and radius \[30\,{\text{cm}}\], and free to rota...

A uniform solid cylinder of mass 5kg5\,{\text{kg}} and radius 30cm30\,{\text{cm}}, and free to rotate about its axis, receives an angular impulse of 3kgm2s13\,{\text{kg}}{{\text{m}}^2}{{\text{s}}^{ - 1}} initially followed by a similar impulse after every 4s{\text{4}}\,{\text{s}}. What is the angular speed of the cylinder 30s{\text{30}}\,{\text{s}} after the initial impulse? The cylinder is at rest initially.
A. 100{\text{100}}
B. 200{\text{200}}
C. 300{\text{300}}
D. 500{\text{500}}

Explanation

Solution

First of all, we will compare the angular impulse and change in angular momentum. From manipulation, will find the angular frequency. We will find the angular acceleration for 4s{\text{4}}\,{\text{s}}. After that we take into account the 30s{\text{30}}\,{\text{s}} and then find the angular speed.

Complete step by step answer:
In the given problem, we are supplied the following data:
Here,
Mass of the body is 5kg5\,{\text{kg}} .
Radius of the cylinder is 30cm30\,{\text{cm}} .
Converting cm{\text{cm}} to m{\text{m}}

1cm=102m 30cm=30×102m 30cm=0.30m 1\,{\text{cm}} = {\text{1}}{{\text{0}}^{ - 2}}{\text{m}} \\\ \Rightarrow 30\,{\text{cm}} = 30 \times {10^{ - 2}}{\text{m}} \\\ \Rightarrow 30\,{\text{cm}} = 0.30\,{\text{m}} \\\

Initial angular speed is zero.
Time after which we need to find out the angular speed is 30s{\text{30}}\,{\text{s}} .
We are asked to find the final angular speed.
To solve this problem, we will first find an expression for angular impulse which is given by change in angular momentum.
Now,
Change in angular momentum is given by:

ΔL=I(ω2ω1) ΔL=12mr2(ω2ω1) \Delta L = I\left( {{\omega _2} - {\omega _1}} \right) \\\ \Rightarrow \Delta L = \dfrac{1}{2}m{r^2}\left( {{\omega _2} - {\omega _1}} \right) \\\

Angular impulse is equal to the change in angular momentum.
So, the equation becomes:
3=12mr2(ω2ω1)3 = \dfrac{1}{2}m{r^2}\left( {{\omega _2} - {\omega _1}} \right) …… (1)
Where,
mm indicates the mass of the body.
rr indicates the radius of the body.
ω2{\omega _2} indicates the final angular frequency.
ω1{\omega _1} indicates the initial angular frequency.
Substituting the required values in the equation (1), we get:

3=12×5×(310)2(ω20) 3=12×5×9100×ω2 ω2=409 3 = \dfrac{1}{2} \times 5 \times {\left( {\dfrac{3}{{10}}} \right)^2}\left( {{\omega _2} - 0} \right) \\\ \Rightarrow 3 = \dfrac{1}{2} \times 5 \times \dfrac{9}{{100}} \times {\omega _2} \\\ \Rightarrow {\omega _2} = \dfrac{{40}}{9} \\\

We know,

ω2=ω1+αt 403=0+α×4 α=103rad/s2 {\omega _2} = {\omega _1} + \alpha t \\\ \Rightarrow \dfrac{{40}}{3} = 0 + \alpha \times 4 \\\ \Rightarrow \alpha = \dfrac{{10}}{3}\,{\text{rad/}}{{\text{s}}^2} \\\

Since, we are asked to find the angular acceleration after 30s30\,{\text{s}}, so do the following operation again, as follows:

ω2=ω1+α×30 ω2=0+103×30 ω2=100rad/s {\omega _2} = {\omega _1} + \alpha \times 30 \\\ \Rightarrow {\omega _2} = 0 + \dfrac{{10}}{3} \times 30 \\\ \Rightarrow {\omega _2} = 100\,\,{\text{rad/s}} \\\

Hence, the angular speed of the cylinder 30s{\text{30}}\,{\text{s}} after the initial impulse is 100rad/s100\,\,{\text{rad/s}} .
The correct option is A.

Note: This problem is based on the concepts of rotation. For a rigid body which is undergoing rotational motion around a fixed axis, the moment of inertia of an object is a computed measure: that is to say, it calculates how difficult it will be to alter the current rotational speed of an object.