A uniform rod of mass M is hinged at its uppe | vedclass

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

Question

Given, mass of the particle $=$ $m$

Mass of the rod $=M$

Let the length of the rod be $L$

Applying conservation of angular momentum about the

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

Given, mass of the particle $=$ $m$

Mass of the rod $=M$

Let the length of the rod be $L$

Applying conservation of angular momentum about the hinge, we get

$m v \frac{L}{2}=I \omega$

$m v \frac{{L}}{2}=\frac{M L^{2}}{3} \omega$

$\Rightarrow \omega=\frac{3 m v}{2 M L}$

Applying principle of conservation of energy, we have

$\frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2}$

$\frac{1}{2} m v^{2}=\frac{1}{2} 	imes \frac{M L^{2}}{3} 	imes\left(\frac{3 m v}{2 M L}ight)^{2}$

$\frac{1}{2} m v^{2}=\frac{1}{2} 	imes \frac{M L^{2}}{3} 	imes \frac{9 m^{2} v^{2}}{4 M^{2} L^{2}}$

$m v^{2}=\frac{3 m^{2} v^{2}}{4 M}$

$\frac{M}{m}=\frac{3}{4}$

$A$ uniform rod of mass $M$ is hinged at its upper end. $A$ particle of mass $m$ moving horizontally strikes the rod at its mid point elastically. If the particle comes to rest after collision find the value of $M/m =?$

Similar Questions

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

The $M.I.$ of a body about the given axis is $1.2\,kg \times m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500\,joule$ an angular acceleration of $25\,rad/sec^2$ must be applied about that axis for a duration of ........ $\sec$.

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be

Write the formula for rotational kinetic energy.

A disc of mass $M$ and radius $R$ rolls in a horizontal surface and then rolls up an inclined plane as shown in the fig. If the velocity of the disc is $v$, the height to which the disc will rise will be..