A body of mass M kg is at the top point of a | vedclass

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

Question

(C) Let the body leave the surface of the hemisphere at point $P$. At this point,the radius vector makes an angle $	heta$ with the vertical.The force

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the body leave the surface of the hemisphere at point $P$. At this point,the radius vector makes an angle $	heta$ with the vertical.The forces acting on the body at point $P$ are the gravitational force $mg$ (acting downwards) and the normal reaction $R$ (acting radially outwards).The component of gravitational force towards the center is $mg \cos 	heta$.The net centripetal force required for circular motion is provided by the difference between the radial component of gravity and the normal reaction:$mg \cos 	heta - R = \frac{mv^2}{r}$When the body leaves the surface,the normal reaction $R$ becomes zero.Therefore,$mg \cos 	heta = \frac{mv^2}{r}$.Substituting the given values $v = 5 \ m/s$,$r = 5 \ m$,and $g = 10 \ m/s^2$:$\cos 	heta = \frac{v^2}{rg} = \frac{5^2}{5 	imes 10} = \frac{25}{50} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = 60^{\circ}$.

Similar Questions

$A$ micelle formed during the cleansing action by soap is

Two circular coils $1$ and $2$ are made from the same wire,but the radius of the $1^{st}$ coil is twice that of the $2^{nd}$ coil. What is the ratio of potential difference in volts that should be applied across them so that the magnetic field at their centres is the same?

In which of the following reactions does hydrogen act as an oxidizing agent?

Which substance is used as a smoke screen in warfare?

Two of the lines represented by the equation $ay^4 + bxy^3 + cx^2y^2 + dx^3y + ex^4 = 0$ will be perpendicular,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module