Question

A ball is tied to the end of a string performs vertical circular motion (complete circle) under the influence of gravity. Choose the correct option(s).

• A. when the string makes an angle 90° with the vertical, the magnitude of tangential acceleration is zero and magnitude of radial acceleration is somewhere between maximum and minimum
• B. when the string makes an angle 90° with the vertical the magnitude of tangential acceleration is maximum and magnitude of radial acceleration is somewhere between maximum and minimum
• C. magnitude of tangential acceleration is never equal to magnitude of radial acceleration
• D. whenever magnitude of radial acceleration has its maximum value, the magnitude of tangential acceleration is maximum

Answer 1

Answer: (B), (C)

The correct options are
B when the string makes an angle 90° with the vertical the magnitude of tangential acceleration is maximum and magnitude of radial acceleration is somewhere between maximum and minimum
C magnitude of tangential acceleration is never equal to magnitude of radial acceleration


By energy conversation:
$\frac{1}{2}m{u}^2$ = $mgl(1-cos\theta )$+$\frac{1}{2}m{v}^2$
Radial acceleration:$a_r=\frac{v^2}{l}=\frac{u^2}{l}-2g+2gcos\theta$
Tangential acceleration: $a_t=gsin\theta$ $a_{rmax}=\frac{u^2}{l};a_{rmin}=\frac{u^2}{l}-4g$
$a_{tmax}=g;a_{tmin}=0$
At $\theta ={90}^o$
$a_r=\frac{u^2}{l}-2g;⟹a_{rmin}<\frac{u^2}{l}-2g<a_{rmax}$
$a_t=g=a_{tmax}$
Let us take:
$a_r=a_t$
$a_r=\frac{u^2}{l}-2g+2gcos\theta =gsin\theta$
On solving further,
$u^2=(-2cos\theta +sin\theta +2)lg$
Since, ball completes full circle, so
$u^2\ge 5gl$
But maximum value of $sin\theta -2cos\theta$ is $(1+2^2{)}^{\frac{1}{2}}$ = $5^{\frac{1}{2}}$
So for no $\theta$
$a_r=a_t$