A solid sphere rolls down an inclined plane a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) When a solid sphere rolls down an inclined plane without slipping,the mechanical energy is conserved. The potential energy at the top is converted

Vedclass · Accepted Answer

$v_1 = \sqrt{\frac{5}{7}} v_2$

Vedclass · Answer

(B) When a solid sphere rolls down an inclined plane without slipping,the mechanical energy is conserved. The potential energy at the top is converted into both translational and rotational kinetic energy at the bottom: $mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}I\omega^2$. For a solid sphere,$I = \frac{2}{5}mr^2$ and $\omega = \frac{v_1}{r}$. Substituting these,$mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_1^2}{r^2}) = \frac{1}{2}mv_1^2 + \frac{1}{5}mv_1^2 = \frac{7}{10}mv_1^2$. Thus,$v_1 = \sqrt{\frac{10}{7}gh}$. When the sphere slides down without friction,there is no rotation,so all potential energy converts to translational kinetic energy: $mgh = \frac{1}{2}mv_2^2$. Thus,$v_2 = \sqrt{2gh}$. Comparing the two,$\frac{v_1}{v_2} = \frac{\sqrt{10/7}}{\sqrt{2}} = \sqrt{\frac{5}{7}}$. Therefore,$v_1 = \sqrt{\frac{5}{7}} v_2$.

$A$ solid sphere rolls down an inclined plane and its velocity at the bottom is $v_1$. Then the same sphere slides down the plane (without friction) and let its velocity at the bottom be $v_2$. Which of the following relations is correct?

Similar Questions

Which of the following (assuming mass and radius are the same) has the maximum percentage of total $K.E.$ in rotational form during pure rolling?

If a cylinder is rolling down an incline with sliding, then which of the following is true?

Which of the following is true about the angular momentum of a cylinder rolling down a slope without slipping?

An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down an inclined plane from rest without slipping has a linear acceleration (where $g$ is the acceleration due to gravity and $\sin 30^{\circ} = 0.5$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module