Question

A point like object of mass M is thrown up from point A as shown in the figure. It slides along the full length of the smooth track ABC (of radius R for part BC). Let $R=1m,AB=2m,M=0.5kg,g=10m/s^2\&OD=3m.$

• A. Minimum speed $V_0$ to slide full length of the track is $\sqrt{60}m/s$
• B. Minimum speed $V_0$ to reach D is $\sqrt{105}m/s$
• C. Normal force on the object by the track at C if it reaches D is 15 N
• D. Normal force on the object by the track at B if it reaches D is 32.5 N

Answer 1

Answer: (B), (D)

Normal force on the object by the track at B if it reaches D is 32.5 N

For the object to slide full length, its normal reaction at C should be greater than zero.
$N_c=m\frac{V_c^2}{R}-mg$
$N_c>0\Rightarrow V_c>\sqrt{Rg}$

By conservation of energy, $\Rightarrow \frac{1}{2}m{V}_0^2=mg\times 3+\frac{1}{2}\times m\times V_c^2$
$\Rightarrow \frac{1}{2}m{V}_0^2=mg\times 3+\frac{1}{2}\times mRg$
$V_0^2=6g+Rg=7g\Rightarrow V_o=\sqrt{70}m/s$
Consider the case where the body falls exactly at D.

By $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$1=0+\frac{1}{2}g{t}^2\Rightarrow t=\frac{1}{\sqrt{5}}$
And,
$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2\Rightarrow 3=V_c\frac{1}{\sqrt{5}}$
So $V_c=3\sqrt{5}$
By conservation of energy,
$\Rightarrow \frac{1}{2}\mathrm{m/}{V}_0^2=\mathrm{m/}g\times 3+\frac{1}{2}\times \mathrm{m/}.(3\sqrt{5}{)}^2$
$V_0^2=60+45\Rightarrow V_0=\sqrt{105}m/s$
Similarly by conservation of energy,
$\Rightarrow V_B=\sqrt{65}$
$N_B=\frac{m{V}_B^2}{R}=\frac{0.5\times 65}{1}=32.5N$