A body is rolling down an inclined plane. If | vedclass

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

Question

(D) Rotational kinetic energy is given by $K_{R} = \frac{1}{2} I \omega^{2} = \frac{1}{2} Mk^{2} \left(\frac{v}{R}\right)^{2}$,where $I = Mk^{2}$ and

Vedclass · Accepted Answer

Solid ball

Vedclass · Answer

(D) Rotational kinetic energy is given by $K_{R} = \frac{1}{2} I \omega^{2} = \frac{1}{2} Mk^{2} \left(\frac{v}{R}ight)^{2}$,where $I = Mk^{2}$ and $v = R \omega$.This simplifies to $K_{R} = \frac{1}{2} Mv^{2} \left(\frac{k^{2}}{R^{2}}ight)$.Translational kinetic energy is given by $K_{T} = \frac{1}{2} Mv^{2}$.According to the problem,$K_{R} = 40\% K_{T}$.Substituting the expressions,we get $\frac{1}{2} Mv^{2} \left(\frac{k^{2}}{R^{2}}ight) = 0.4 	imes \frac{1}{2} Mv^{2}$.This implies $\frac{k^{2}}{R^{2}} = 0.4 = \frac{2}{5}$.For a solid sphere (solid ball),the moment of inertia is $I = \frac{2}{5} MR^{2}$,so $k^{2} = \frac{2}{5} R^{2}$,which means $\frac{k^{2}}{R^{2}} = \frac{2}{5}$.Therefore,the body is a solid ball.

$A$ body is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a

Similar Questions

$A$ ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\frac{7}{x}$ where $x$ is . . . . . . .

$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. Identify the correct statement$(s)$.

$A$ small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

Starting from rest,at the same time,a ring,a coin (disc),and a solid ball of the same mass roll down an incline without slipping. The ratio of their translational kinetic energies at the bottom will be:

$A$ ball of radius $11 \ cm$ and mass $8 \ kg$ rolls from rest down a ramp of length $2 \ m$. The ramp is inclined at $35^{\circ}$ to the horizontal. When the ball reaches the bottom,its velocity is .......... $m/s$. $(\sin 35^{\circ} = 0.57)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module