Question

Question: Figure shows the potential energy U (x) of a solid ball that can roll along an xaxis. The scale on the Uaxis is set by U_s =100j. The ball is uniform, rolls smoothly, and has a mass of 0.400 kgIt is released at x = 7.0 mheaded in the negative direction of the xaxis with a mechanical energy of 75 J (a) If the ball can reach x = 0 m, what is its speed there, and if it cannot, what is its turning point? Suppose, instead, it is headed in the positive direction of the xaxis when it is released at x = 7.0 m with 75J(b) If the ball can reach x = 13m, what is its speed there, and if it cannot, what is its turning point?

Short answer

Answer

1. The ball does not reach x = 0 m. The turning point of the ball is at x = 2 m .
2. The ball reaches the pointx = 13 m . Its velocity there is 7.3 m.

Step-by-step solution

Given

1. The mass of the ball is M = 0.4 KJ
2. The mechanical energy of the ball at x =7 is 75J

To understand the solution

Using the law of conservation of energy, we can find the P.E at the turning point. Then observing the graph, we can find that point. From that, we can conclude whether the ball reaches x = 0 m.

For part b, by observing the graph, we can conclude whether there is a turning point at x = 13 m. Then we can use the law of conservation of energy to find the total energy at point x = 13 m.

From the graph, we will get the P.E at that point. Then substituting it in the total energy, we will get the total K.E of the ball at x = 13 m. Then substituting$\omega$in terms of v and the M.I of the ball (sphere) we will get the value for velocity of the ball at the point x = 13 m.

The law of conservation of energy is,

U + K = Constant

The angular velocity is given as-

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

(a) If the ball can reach , Calculate its speed there, and if it cannot, calculate its turning point

Let the turning point be A.

The energy at x = 7.0 m is 75 J. So, according to the conservation of energy, the energy at point A is,

$$\begin{gathered} E=K+U \\ =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+U \\ =75\text{J} \end{gathered}$$

At the turning point, velocity and angular velocity is zero. So, the above equation becomes,

. U = 75 J

From the given graph, we can infer that it is at point X = 2.00m.

Therefore, we can conclude that the ball does not reach the origin.

(b) If the ball can reach , calculate its speed there, and if it cannot, calculate is its turning point

From part (a), we can say that the turning point should have P.E (U) equal to 75J, and from the graph, we get there is no such point after x = 7.0 m. So, it is not possible to have turning point at x = 13 m .

By energy conservation, we can write that,

$$\begin{gathered} E=K_{translational}+K_{rotational}+U \\ =75\text{J} \\ \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+U=75\text{J} \end{gathered}$$

From the given graph, we can infer that,

U = 60 J

So,

$$\begin{gathered} \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+60\text{J}=75\text{J} \\ \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=15\text{J} \end{gathered}$$

But, we know that rotational inertia the ball is,

$I=\frac{2}{5}M{R}^2$

The relation between v and ω is,

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

So,

$$\begin{gathered} \frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}M{R}^2(v^2/R^2)=15\text{J} \\ \frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=15\text{J} \\ \frac{7}{10}m{v}^2=15\text{J} \\ 0.7\left(0.4\text{kg}\right){\text{v}}^2=15\text{J} \\ v=7.3m/s \end{gathered}$$

Therefore, the velocity of the ball at point x = 13 m is 7.3 m/s.