A particle is rotating in a circle of radius $1\,m$ with constant speed $4\,m / s$. In time $1\,s$, match the following (in $SI$ units) columns.
Colum $I$ | Colum $II$ |
$(A)$ Displacement | $(p)$ $8 \sin 2$ |
$(B)$ Distance | $(q)$ $4$ |
$(C)$ Average velocity | $(r)$ $2 \sin 2$ |
$(D)$ Average acceleration | $(s)$ $4 \sin 2$ |
$( A \rightarrow r , B \rightarrow q , C \rightarrow r , D \rightarrow p )$
$( A \rightarrow p , B \rightarrow q , C \rightarrow r , D \rightarrow p )$
$( A \rightarrow r , B \rightarrow s , C \rightarrow r , D \rightarrow p )$
$( A \rightarrow p , B \rightarrow q , C \rightarrow r , D \rightarrow s)$
A particle moving in a circle of radius $R$ with uniform speed takes time $\mathrm{T}$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4 R$. The angle of projection $\theta$ is then given by :
The acceleration vector of a particle in uniform circular motion averaged over the cycle is a null vector. This statement is
What is uniform circular motion ? By using proper figure, obtain equation of acceleration ${a_c}\, = \,\frac{{{v^2}}}{r}$ for uniform circular motion. Show that its direction is towards centre.
A point $P$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of '$P$' is such that it sweeps out a length $s = t^3+5$, where s is in metres and $t$ is in seconds. The radius of the path is $20\ m$. The acceleration of '$P$' when $t = 2\ s$ is nearly .......... $m/s^2$
A small block slides down from rest at point $A$ on the surface of a smooth cylinder, as shown. At point $B$, the block falls off (leaves) the cylinder. The equation relating the angles $\theta_1$ and $\theta_2$ is given by