A cyclist starts from centre 0 of a circular park of radius $1\, km$ and, moves along the path $OPRQO$ as shown in figure.
If he maintains constant speed of $10\, ms^{-1}$, what is his acceleration at point $R$ in magnitude and direction ?
Here, object performs circular motion. Hence, its acceleration is called centripetal acceleration.
Hence, acceleration at $\mathrm{R}$ is,
$a=\frac{v^{2}}{r}=\frac{(10)^{2}}{1 \mathrm{~km}}=\frac{100}{10^{3}}$
$=0.1 \mathrm{~m} / \mathrm{s}^{2}$ along $\mathrm{RO}$ (towards centre)
A particle moving with uniform speed in a circular path maintains:
For a particle in uniform circular motion, the acceleration $\vec a$ at a point $P(R,\theta)$ on the circle of radius $R$ is (Here $\theta$ is measured from the $x-$ axis)
A particle is going with constant speed along a uniform helical and spiral path separately as shown in figure
Two bodies of mass $10 \,kg$ and $5 \,kg$ moving in concentric orbits of radii $R$ and $r$ such that their periods are the same. Then the ratio between their centripetal acceleration is
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