Question 1:
moderateA sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be
\[ mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}= \frac{1}{2}mv^{2}+ \frac{1}{2}(\frac{2}{5}mR^{2})\frac{v^{2}}{R^{2}} \]
Solving we get,
\[ v=Â \sqrt[]{\frac{10}{7}gh} \]
Question 2:
moderateA body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is
Given, rotational kinetic energy is 40% of total energy. so,
\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]
Solving ,
\[ I = \frac{2}{5}mR^{2} \]
Object is Solid Sphere.
Question 3:
moderateA disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?
(2016 – I)
Acceleration of a rolling body on an inclined plane is given by $$a = \frac{g sin\theta}{1 + I/(mR^2)}$$. Since the acceleration is independent of mass and depends only on the geometry (moment of inertia factor), the sphere has a smaller inertia factor than the disk, meaning the sphere has a greater acceleration and reaches the bottom first.
Question 4:
moderateThe ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle $theta$ without slipping and slipping down the incline without rolling is:
(2014)
Acceleration without slipping is $a_1 = \frac{g sin\theta}{1 + I/(mR^2)} = \frac{5}{7}g sin\theta$. Acceleration with pure slipping is $a_2 = g sin\theta$. The ratio $a_1/a_2$ is $5/7$.
Question 5:
moderateSmall object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is:
(2013)
Using energy conservation, initial kinetic energy equals potential energy at max height: $\frac{1}{2}mv^2 \left(1 + \frac{I}{mR^2}\right) = mgH$. Substituting $H = \frac{3v^2}{4g}$, we get $1 + \frac{I}{mR^2} = 2$, which gives $\frac{I}{mR^2} = 1$. This corresponds to a ring.
Question 6:
easyA solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
(2010 Mains)
The acceleration of a rolling body depends on its moment of inertia ratio $I/mR^2$. The solid cylinder has a smaller moment of inertia ratio ($1/2$) compared to the hollow cylinder ($1$), giving it a higher acceleration and causing it to reach the bottom first.
Question 7:
moderateFor a hollow cylinder & a solid cylinder rolling without slipping on an inclined plane, then which of these reaches earlier on the ground:
(2000)
Acceleration of a rolling body is given by $$a = \frac{g \sin\theta}{1 + I/MR^2}$$. Since the solid cylinder has a smaller moment of inertia ratio than the hollow cylinder, its acceleration is greater, so it reaches the bottom first.
Question 8:
moderateA solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on the inclined plane, then:
(1993)
The acceleration on an inclined plane is inversely proportional to $1 + I/MR^2$. Solid sphere has the lowest moment of inertia coefficient ($2/5$), giving it maximum acceleration and shortest time to reach the bottom.
Question 9:
moderateThe speed of a homogenous solid sphere after rolling down an inclined plane of vertical height $h$ from rest without sliding is:
(1992)
Using conservation of mechanical energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For a solid sphere ($I = \frac{2}{5}MR^2$), solving yields $v = \sqrt{\frac{10}{7}gh}$.