From an inclined plane,a ring,a disc,a sphere | vedclass

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

Question

(C) The acceleration $a$ of a body rolling down an inclined plane is given by the formula: $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$,where $I$ is

Vedclass · Accepted Answer

Solid Sphere

Vedclass · Answer

(C) The acceleration $a$ of a body rolling down an inclined plane is given by the formula: $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia,$M$ is the mass,and $R$ is the radius.For a given inclined plane,$g$ and $	heta$ are constant. Therefore,the acceleration $a$ is inversely proportional to the factor $(1 + \frac{I}{MR^2})$.The body with the smallest value of $\frac{I}{MR^2}$ will have the largest acceleration and will reach the base first.- For a ring: $I = MR^2$,so $\frac{I}{MR^2} = 1$.- For a disc: $I = \frac{1}{2}MR^2$,so $\frac{I}{MR^2} = 0.5$.- For a solid sphere: $I = \frac{2}{5}MR^2$,so $\frac{I}{MR^2} = 0.4$.- For a cylinder: $I = \frac{1}{2}MR^2$,so $\frac{I}{MR^2} = 0.5$.Comparing the values,the solid sphere has the smallest value $(0.4)$.Thus,the solid sphere has the highest acceleration and reaches the base first.

From an inclined plane,a ring,a disc,a sphere,and a cylinder of equal radius are rolled without slipping. Which one reaches the base first?

Similar Questions

$A$ ring of radius $4a$ is rigidly fixed in a vertical position on a table. $A$ small disc of mass $m$ and radius $a$ is released as shown in the figure. When the disc rolls down,without slipping,to the lowest point of the ring,then its speed will be

$A$ solid cylinder and a disc of same radii are allowed to roll down a rough inclined plane from the top of a plane. The ratio of their times taken to reach the bottom of the inclined plane is

$A$ disc of radius $20 \, cm$ and mass $0.5 \, kg$ is rolling on an inclined plane with an angle of inclination $45^{\circ}$. Find the friction force required for the disc to perform pure rolling.

$A$ solid cylinder of diameter $30 \ cm$ is rolled down an inclined plane from a height of $2 \ m$. If there is no energy loss due to friction,the angular velocity at the base of the plane is ....... $rad/s$. (Take $g = 10 \ m/s^2$)

$A$ solid sphere rolls without slipping on an inclined plane at an angle $\theta$. The ratio of total kinetic energy to its rotational kinetic energy is

Vedclass Test Series

Exam Paper Generator

Online Exam Module