A body kept on a smooth horizontal surface is | vedclass

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

Question

(A) For translational motion of the center of mass $(C.M.)$:$F + f = ma$  (where $f$ is the friction force)For rotational motion about the $C.M.$:$R(F

Vedclass · Accepted Answer

thin pipe

Vedclass · Answer

(A) For translational motion of the center of mass $(C.M.)$:$F + f = ma$  (where $f$ is the friction force)For rotational motion about the $C.M.$:$R(F - f) = I\alpha$For pure rolling,the condition is $a = R\alpha$,which implies $\alpha = a/R$.Substituting $\alpha$ in the rotational equation:$R(F - f) = I(a/R) \implies F - f = \frac{I}{R^2} a$From the translational equation,$a = \frac{F + f}{m}$. Substituting this:$F - f = \frac{I}{R^2} \left( \frac{F + f}{m} \right)$$mR^2(F - f) = I(F + f)$$F(mR^2 - I) = f(mR^2 + I)$$f = F \frac{mR^2 - I}{mR^2 + I}$For pure rolling to occur on a smooth surface without needing external friction,we require $f = 0$.This implies $mR^2 - I = 0$,or $I = mR^2$.Among the given options,a thin pipe (or ring) has a moment of inertia $I = mR^2$ about its central axis.

$A$ body kept on a smooth horizontal surface is pulled by a constant horizontal force applied at the top point of the body. If the body rolls purely on the surface,its shape can be:

Similar Questions

From a solid sphere of mass $M$ and radius $R$,a cube of maximum possible volume is cut. The moment of inertia of the cube about an axis passing through its center and perpendicular to one of its faces is

$A$ thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of:

$A$ particle starts from the point $(0\,m, 8\,m)$ and moves with a uniform velocity of $3\, \hat{i} \,m/s$. After $5\,s$,the angular velocity of the particle about the origin will be:

$A$ uniform disc is acted upon by two equal forces of magnitude $F$. One of them acts tangentially to the disc,while the other acts at the central point of the disc. The friction between the disc surface and the ground surface is $nF$. If $r$ is the radius of the disc,then the value of $n$ would be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module