Question:
A small body of mass $m$ slides without friction from the top of a hemisphere of radius $r$. At what height will the body be detached from the centre of the hemisphere?
A small body of mass $m$ slides without friction from the top of a hemisphere of radius $r$. At what height will the body be detached from the centre of the hemisphere?
Updated On: Jul 5, 2022
- $\frac{r}{3}$
- $\frac{r}{2}$
- $\frac{2}{3}r$
- 2r
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The Correct Option is C
Solution and Explanation
Body will be detached if
$\frac{m \nu^{2}}{r}=m g \cos \theta$
As $\nu^{2}-0^{2}=2 g d$
$\nu=\sqrt{2 g d}$
and $\cos \theta=\frac{h}{r}$
So, from E (i),
$\frac{m \times 2 g d}{r}=m g \frac{h}{r}$
or $h=2 d$ or $h=2(r-h)$
ie, $h=(2 / 3) r$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration