Home
Class 12
PHYSICS
A disc is rolling on a surface without s...

A disc is rolling on a surface without slipping. What is the ratio of its translational to rotational kinetic energies ?

A

`5 : 2`

B

`2: 1`

C

`3 : 2`

D

`2 : 3`

Text Solution

AI Generated Solution

The correct Answer is:
To find the ratio of translational to rotational kinetic energies for a disc rolling without slipping, we can follow these steps: ### Step 1: Define the Kinetic Energies The translational kinetic energy (TKE) of the disc is given by the formula: \[ \text{TKE} = \frac{1}{2} m v^2 \] where \( m \) is the mass of the disc and \( v \) is the velocity of the center of mass. The rotational kinetic energy (RKE) of the disc is given by the formula: \[ \text{RKE} = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. ### Step 2: Determine the Moment of Inertia For a solid disc, the moment of inertia about its center is: \[ I = \frac{1}{2} m r^2 \] where \( r \) is the radius of the disc. ### Step 3: Relate Linear and Angular Velocity For a disc rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is: \[ v = r \omega \] From this, we can express \( \omega \) in terms of \( v \): \[ \omega = \frac{v}{r} \] ### Step 4: Substitute \( \omega \) in RKE Now, substituting \( \omega \) in the expression for rotational kinetic energy: \[ \text{RKE} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \] This simplifies to: \[ \text{RKE} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{4} m v^2 \] ### Step 5: Calculate the Ratio of TKE to RKE Now we can find the ratio of translational kinetic energy to rotational kinetic energy: \[ \text{Ratio} = \frac{\text{TKE}}{\text{RKE}} = \frac{\frac{1}{2} m v^2}{\frac{1}{4} m v^2} \] The \( m v^2 \) terms cancel out: \[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{1}{4}} = \frac{1/2}{1/4} = 2 \] ### Conclusion Thus, the ratio of translational to rotational kinetic energies for a disc rolling without slipping is: \[ \text{Ratio} = 2:1 \]
Promotional Banner

Similar Questions

Explore conceptually related problems

A solid sphere rolls on horizontal surface without slipping. What is the ratio of its rotational to translation kinetic energy.

If a body is rolling on a surface without slipping such that its kinetic energy of translation is equal to kinetic energy of rotation then it is a

A ring is rolling without slipping. The ratio of translational kinetic energy to rotational kinetic energy is

A disc is rolling without slipping. The ratio of its rotational kinetic energy and translational kinetic energy would be -

A hollow sphere is rolling without slipping on a rough surface. The ratio of translational kinetic energy to rotational kinetic energy is

If a rigid body rolls on a surface without slipping, then:

When a sphere rolls without slipping the ratio of its kinetic energy of translation to its total kinetic energy is.

The radius of a wheel is R and its radius of gyration about its axis passing through its center and perpendicualr to its plane is K. If the wheel is roling without slipping. Then the ratio of tis rotational kinetic energy to its translational kinetic energy is

When a solid cylinder rolls without slipping the ratio of kinetic energy of translation to its total kinetic energy is

The moment of inertia of a solid cylinder about its axis is given by (1//2)MR^(2) . If this cylinder rolls without slipping the ratio of its rotational kinetic energy to its translational kinetic energy is -