Question

A block is projected with a speed $v_0$ such that it strikes the point of projection after describing the path as shown by the dotted line. If friction exists for the path of length d and the vertical circular path is smooth assuming $mu$ = coefficient of friction,
a. Find $v_0$.
b. What is the minimum value of $v_0$?

• A. a. $\sqrt{\frac{d^{2}g}{4R}+2g[\mu d+2R]}$; b. $\sqrt{5gR}$
• B. a. $\sqrt{\frac{d^{3}g}{4R}+3g[\mu d+2R]}$; b. $\sqrt{5gR}$
• C. a. $\sqrt{\frac{d^{3}g}{4R}+3g[\mu d+3R]}$; b. $\sqrt{5gR}$
• D. a. $\sqrt{\frac{d^{3}g}{4R}+4g[\mu d+2R]}$; b. $\sqrt{5gR}$

Answer 1

The correct option is A a. $\sqrt{\frac{d^{2}g}{4R}+2g[\mu d+2R]}$; b. $\sqrt{5gR}$

a) From $C$ to $A$, time taken

$t=\sqrt{\frac{2\left[2R\right]}{g}}=\sqrt{\frac{4R}{g}}$

$v_2=\frac{d}{t}=d\sqrt{\frac{g}{4R}}$

From $B$ to $C,\frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}_2^2+mg\left(2R\right)$$

$\Rightarrow v_1^2=v_2^2+4gR=\frac{d^{2}g}{4R}+4g^R$

From $A$ to $B:v_1^2=v_0^2-2\mu g$

$\Rightarrow v_0^2=v_1^2+2\mu gd$

$\Rightarrow v_0^2=\frac{d^{2}g}{4R}+2g[\mu d+2R]$

b) For minimum $v_0$

$V_1=\sqrt{5gR}$