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Question: A solid cylinder of mass \(20kg\) rotates about its axis with angular speed \(100rad{s^{ - 1}}\) . T...

A solid cylinder of mass 20kg20kg rotates about its axis with angular speed 100rads1100rad{s^{ - 1}} . The radius of the cylinder is 0.25m0.25m . What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Explanation

Solution

We have to go with the formula of moment of inertia I=mr22I = \dfrac{{m{r^2}}}{2} . Then calculate kinetic energy by formula which is equal to E=Iω22E = \dfrac{{I{\omega ^2}}}{2} and angular momentum by formula which is equal to L=IωL = I\omega .

Formula Used:
Moment of Inertia for a cylinder, I=mr22I = \dfrac{{m{r^2}}}{2}
Where mm is the mass of the body, rr is the radius of the cylinder.
Kinetic Energy, E=Iω22E = \dfrac{{I{\omega ^2}}}{2}
Where, II is the moment of inertia of the body, ω\omega is the angular speed of the cylinder.
Angular Momentum, L=IωL = I\omega
Where, II is the moment of inertia of the body, ω\omega is the angular speed of the cylinder.

Complete step by step solution:
Firstly, we will note the given value and check if they are in the Si unit or not. If not, we will convert them in their respective SI units.
Mass, mm of the body is equal to 20kg20kg (this is in SI units)
Angular speed, (ω)\left( \omega \right) is equal to 100rads1100rad{s^{ - 1}} (this is also in Si unit)
Radius of cylinder is equal to 0.25m0.25m (also in SI unit)
Now calculate the moment of inertia which is defined as force with respect to a specific rotational axis. The moment of inertia is the product of mass and the square of the distance from the axis. Therefore,
It is represented by
I=mr22I = \dfrac{{m{r^2}}}{2}
Insert above given values in the above equation, we get
12×20×(0.25)2\Rightarrow \dfrac{1}{2} \times 20 \times {\left( {0.25} \right)^2}
10×0.0625\Rightarrow 10 \times 0.0625
So, we have moment of inertia which is equal to
0.625kgm2\Rightarrow 0.625kg{m^2}
After this, find the kinetic energy using the below mentioned formula:
E=Iω22E = \dfrac{{I{\omega ^2}}}{2}
Insert above given values
E=12×0.625×(1002)E = \dfrac{1}{2} \times 0.625 \times \left( {{{100}^2}} \right)
Therefore, kinetic energy is equal to
E=3125J\Rightarrow E = 3125J
Now, calculate the angular momentum.
L=IωL = I\omega
L=0.625×100\Rightarrow L = 0.625 \times 100

So angular momentum is L=62.5JsL = 62.5Js.

Note: Always remember that we have to use SI units of all the quantities otherwise the answer is miscalculated. Also, here all formulas used for rotating objects so don’t go for linear formulas of quantities.