Question

The following figure shows a small spherical ball of mass $m$ rolling down the loop track. The ball is released on the linear portion at a vertical height $H$ from the lowest point. The circular part shown has a radius $R$ .
$(a)$ Find the kinetic energy of the ball when it is at a point $A$ where the radius makes an angle $θ$ with the horizontal.

$(b)$ Find the radial and the tangential accelerations of the center when the ball is at $A$ .

(c)

Find the normal force and the frictional force acting on the ball if

H = 60 cm, R = 10 cm, θ = 0 °

and

m = 70 g

.

Accepted Answer

Let the velocity and angular velocity of the ball at point $A$ be $v$ and $ω$ respectively.

Total kinetic energy at point $A = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

Total potential energy at point $A$

$= m g (R + R s i n θ)$

On applying the law of conservation of energy, we have,

Total energy at initial point = Total energy at $A$

Therefore, we get,

$0 + m g H = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} + m g R (1 + s i n θ)$

$\Rightarrow m g H - m g R (1 + s i n θ) = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

$\Rightarrow \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g (H - R - R s i n θ) . . . (i)$

Total K.E at $A = m g (H - R - R s i n θ)$

Now, find the acceleration component putting,

$I = \frac{2}{5} m R^{2}$ and $ω = \frac{V}{R}$

In equation $(i)$ , we get,

$\frac{7}{10} m v^{2} = m g (H - R - R s i n θ)$

$v^{2} = \frac{10}{7} g (H - R - R s i n θ) . . . (ii)$

Radial acceleration,

$a_{r} = \frac{v^{2}}{R} = \frac{10}{7} g (\frac{H - R - R s i n θ}{R}) = \frac{10}{7} g (\frac{H}{R} - 1 - s i n θ)$

For tangential acceleration,

Differentiating equation $(ii)$ w.r.t $t$

$2 v \frac{d v}{d t} = - (\frac{10}{7}) g R c o s θ \frac{d θ}{d t}$

$\Rightarrow ω R \frac{d v}{d t} = (- \frac{5}{7}) g R c o s θ \frac{d θ}{d t} \Rightarrow \frac{d v}{d t} = - (\frac{5}{7}) g c o s θ \Rightarrow a_{t} = - (\frac{5}{7}) g c o s θ .$

At $θ = 0 °$ , from free body diagram w.r.t. ball, we will consider centrifugal force, we have,

Normal force $= N = m a_{r}$

$N = m \times \frac{10}{7} g \frac{(H - R - R s i n θ)}{R}$

$= \frac{70}{1000} \times (\frac{10}{7}) \times 9.8 (\frac{0.6 - 0.1}{0.1}) = 4.9 N$

At $θ = 0 °$ , from the free body diagram, $f_{r} =$ frictional force

$m g - f_{r} = m a_{t} f_{r} = m g - m a_{t} f_{r} = m (g - a_{t})$

$= m (9.8 - \frac{5}{7} \times 9.8)$

$= 0.07 (9.8 - \frac{5}{7} \times 9.8) = 0.196 N$

H C Verma Solutions for Chapter: Rotational Mechanics, Exercise 4: EXERCISES

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