Question

A wheel of mass 5kg and radius 0.40m is rolling on a road without sliding with angular velocity 10rad s-1.
The moment of inertia of the wheel about the axis of rotation is 0.65kgm2. The percentage of kinetic
energy of rotation to the total kinetic energy of the wheel is:
1) 22.4%
2) 11.2%
3) 88.8%
4) 44.8%​

Answer 1

Answer:

2)11.2 is the correct answer

Answer 2

Answer:

The percentage of Kinetic Energy of rotation to the Kinetic Energy of the wheel.

Explanation:

Given:

Mass = 5 Kg

Radius = 0.40 m

Angular velocity ,ω = 10 rad $s^{-1}$

Axis of rotation ,I = 0.65 kgm²

Concept:

Linear Velocity-It is defined as the changing to displacement with time  when the object moves along a straight path is known to be linear velocity and it is written as;

v = rω    .....(1)

Translation Kinetic Energy:  when we applied work to the object it accelerate from rest position by means of velocity . It is written as;

K.E.=$\frac{1}{2}mv^{2}$   ....(2)

Rotational Kinetic energy - It is defined as the energy due to the rotation of the object. It is written as;

K.E =$\frac{1}{2}I$ω²    ....(3)

Calculation:

Linear velocity ,v = Rω

v = 0.40×10

v = 4 m$s^{-1}$

Translation Kinetic Energy ,K.E.=$\frac{1}{2} mv^{2}$

                                                    =$\frac{1}{2}$×5×4²

                                                    =$\frac{1}{2}$×5×16

                                                   = 40 J

Rotational Kinetic Energy, K.E = $\frac{1}{2}$×0.65×100

                                                 = 32.5 J

Total Kinetic energy =Translational K.E. + Rotational K.E.

                                  = 40+32.5

                                 = 72.5 J

Percentage of kinetic energy is equal to =$\frac{Rotational }{Total }$×100

                                                                    =$\frac{32.5}{72.5}$×100

                                                                    = 44.8 %

Result:

Hence option 4 is correct.