Physics - Circular Motion Question with Solution | TestHub

The body moves on a circular arc of radius $\ell$ centred at the attachment point $P$.

Resolving along the thread at angle $\phi$,

$$T-mg\text{sin}\alpha \text{sin}\phi =\frac{m{v}^2}{\ell }\dots (i)$$

and along the tangent to the path (on the slope),

$$mg\text{sin}\alpha \text{cos}\phi -\mu mg\text{cos}\alpha =m\frac{dv}{dt}\dots (ii)$$

Applying the work--energy theorem between release ($\phi =0$) and angle $\phi$,

$$(mg\text{sin}\alpha )\ell \text{sin}\phi -(\mu mg\text{cos}\alpha )\ell \phi =\frac{1}{2}m{v}^2\dots (iii)$$

From (i) and (iii),

$$\Rightarrow T-mg\text{sin}\alpha \text{sin}\phi =2mg\text{sin}\alpha \text{sin}\phi -2(\mu mg\text{cos}\alpha )\phi$$

$$\Rightarrow T=mg(3\text{sin}\alpha \text{sin}\phi -2\mu \phi \text{cos}\alpha )$$

so (A) is correct. The tension is maximum when

$$\frac{dT}{d\phi }=0=3\text{sin}\alpha \text{cos}\phi -2\mu \text{cos}\alpha \Rightarrow \phi ={\text{cos}}^{-1}\left(\frac{2\mu }{3\text{tan}\alpha }\right)={\text{cos}}^{-1}\left(\frac{2\times 0.2}{3\text{tan}30^o}\right)\approx {76.6}^o<90^o$$

so (C) is correct and (B) is false. The speed is maximum when the tangential acceleration vanishes, i.e. from (ii),

$$mg\text{sin}\alpha \text{cos}\phi =\mu mg\text{cos}\alpha \Rightarrow \phi ={\text{cos}}^{-1}(\mu \text{cot}\alpha )$$

so (D) is also correct.