A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution after the revolution has started then the tangential acceleration is

Question

A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution after the revolution has started then the tangential acceleration is (A) (v2/8 π r) (B) (v2/6 π r) (C) (v2/4 π r) (D) (v2/10 π r)

Tardigrade Admin · Accepted Answer

Using v2-u2=2 a s and u=u1 and s=4 π r ∴ 2 a s=v2 ⇒ (v2/2 s)=a=(v2/2 × 4 π r)=(v2/8 π r)

Q. A particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of second revolution after the revolution has started then the tangential acceleration is

$\frac{v ^{2}}{8 π r}$

$\frac{v ^{2}}{6 π r}$

$\frac{v ^{2}}{4 π r}$

$\frac{v ^{2}}{10 π r}$

Using $v^{2} - u^{2} = 2 a s$ and $u = u_{1}$ and $s = 4 π r$
$∴ 2 a s = v^{2} \Rightarrow \frac{v ^{2}}{2 s} = a = \frac{v ^{2}}{2 \times 4 π r} = \frac{v ^{2}}{8 π r}$

Q. A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution after the revolution has started then the tangential acceleration is

8πrv2​

6πrv2​

4πrv2​

10πrv2​