Question

Question: In Figure, a solid brass ball of mass 0.280 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radiusr=14.o cm and the ball has radius r<<R(a) What is hif the ball is on the verge of leaving the track when it reaches the top of the loop? If the ball is released at height, h = 6.00R (b) What is the magnitude of the horizontal force component acting on the ball at point Q? (c) What is the direction of the horizontal force component acting on the ball at point Q?

Short answer

Answer

1. The height h if the ball is on the verge of leaving the track when it reaches the top of the loop is,0.378 m.
2. The magnitude of the horizontal force component acting on the ball at point Q is.$1.96\times {10}^{-2}\text{N}$
3. The direction of the horizontal force component acting on the ball at point Q is towards the center of the loop.

Step-by-step solution

Given

1. The mass of the solid brass ball,$\text{m}=0.280\text{kg}.$
2. The radius of the circular loop is,$\text{R}=0.14\text{m}.$

To understand the concept

The law of conservation of energy states that the mechanical energy for an isolated system remains conserved, in the absence of dissipative forces.

The law conservation of energy is given as-

$\mathbf{U}\mathbf{+}{\mathbf{K}}_{\mathbf{tran}}\mathbf{+}{\mathbf{K}}_{\mathbf{rot}}\mathbf{=}\mathbf{}\mathbf{constant}$

The angular velocity is given as-

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{v}}{\mathbf{R}}$

The moment of inertial for a solid sphere is given as-

${\mathbf{I}}_{\mathbf{solidsphere}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{5}}{\mathbf{mr}}^{\mathbf{2}}$

(a) Calculate h if the ball is on the verge of leaving the track when it reaches the top of the loop  

At height h , the ball is at rest, so the total mechanical energy is just potential energy. If we consider potential energy at the ground, to be zero, then at ground the total mechanical energy is kinetic energy. Also, at the top of the loop the potential energy will be 2mgr . According to law of conservation of energy, the total mechanical energy should be conserved. So,

Total energy of the ball at height h = Total energy of the system at height 2R

From figure we can write it as,

$$\begin{gathered} K_h+U_h=K_{2R}+U_{2R} \\ 0+mgh=\left(\frac{1}{2}m{v}^2\right)+\left(\frac{1}{2}I{\omega }^2\right)+mg\left(2R\right) ······································\left(1\right)\backslash \mathrm{hfill} \end{gathered}$$

Height of the loop would be equal to its diameter, so we can write it as .

But,

$$\begin{gathered} \left(\frac{1}{2}m{v}^2\right)+\left(\frac{1}{2}I{\omega }^2\right)=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\left(\frac{v^2}{r^2}\right) \\ \left(\frac{1}{2}m{v}^2\right)+\left(\frac{1}{2}I{\omega }^2\right)=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2 \\ \left(\frac{1}{2}m{v}^2\right)+\left(\frac{1}{2}I{\omega }^2\right)=\frac{7}{10}m{v}^2 \end{gathered}$$

Substituting in equation (1)

$mgh=\left(\frac{7}{10}m{v}^2\right)+mg2R··································\left(1a\right)$

If the ball is on the verge of leaving the track when it reaches the top of the loop then Normal force acting on it is negligible. Also it is given that r>>R. So the Newton’s second law for this situation becomes,

$mg=\frac{m{v}^2}{R}$.

$v^2=gR$----(2)

Putting equation (2) in equation (1a) we get,

$$\begin{gathered} mgh=\left(\frac{7}{10}mgR\right)+mg\left(2R\right) \\ h=\left(\frac{7}{10}R\right)+\left(2R\right) \\ h=\frac{27}{10}R \end{gathered}$$

For the given values-

$$\begin{gathered} h=\left(2.7\right)\left(0.14\text{m}\right) \\ h=0.378\text{m} \end{gathered}$$

Therefore, the height h if the ball is on the verge of leaving the track when it reaches the top of the loop is0.378 m.

(b) Calculate the magnitude of the horizontal force component acting on the ball at point Q  

If the ball is released at h=6R and it reaches at point Q then equation (1) becomes,

$$\begin{gathered} mg\left(6R\right)=\left(\frac{7}{10}m{v}^2\right)+mg\left(R\right) \\ 6gR=\left(\frac{7}{10}{v}^2\right)+gR \\ {v}^2=\frac{50gR}{7} \end{gathered}$$

If the ball is released at h = 6R and it reaches at point Q then by applying Newton’s second law, we get-

$$\begin{gathered} N=\frac{m{v}^2}{R-r} \\ N=\frac{m50gR}{7\left(R-r\right)} \\ Forr<<R \\ \Rightarrow N=\frac{m50gR}{7R} \\ N=\frac{50mg}{7} \end{gathered}$$

For the given values-

$N=\frac{50(2.8\times {10}^{-4}\text{kg})(9.8m/s^2)}{7}$

$N=1.96\times {10}^{-2}\text{N}$

Therefore, the magnitude of the horizontal force component acting on the ball at point Q is$N=1.96\times {10}^{-2}\text{N}$

(c) Calculate the direction of the horizontal force component acting on the ball at point Q

For the ball to go around the circular orbit, it needs to have the centripetal force acting on it. The weight of the ball is acting vertically downwards, so this force would not provide any centripetal force at point Q. There is normal reaction on the ball from the surface of the loop at point Q. It is the only force acting on the object in the horizontal direction which would cause an object to undergo centripetal motion. The directionof the centripetal force acting on the ball would always be pointing towards the center. Hence the force at point Q is towards the center of the loop.