Chapter 8, Problem 70QAP

To determine

Normal force acting on the billiard ball at point $B$

Answer to Problem 70QAP

  $F_B=1.143N$

Explanation of Solution

Given info:

Billiard ball of,

  Radius $=r_0=2.50cm$

  Mass $=m_o=160.0g$

Translational speed of billiard ball at $A=$ $2.00m{s}^{-1}$ $=v_0$

Point B is at the top of a hill that has a radius of curvature of $60cm$

  $(r)$

Billiard ball rolls without slipping down the track.

Formula used:

Let's name normal force acting on the ball at point $B$

As $F_B$.

Let's name the angular velocity of billiard ball at the top of the track as $\omega$.

Let's name the linear speed of billiard ball at the top of the hill as $v_s$

Let's name the moment of inertia of billiard ball as $I_s$.

Let's name the vertical distance from potential energy zero level $=h$ $=-10cm$

  $g=10m{s}^{-2}$.

Conservation of mechanical energy:

  $K_i+U_i=K_f+U_f$

Kinetic energy for an object that undergoes both translation and rotation:

  $K=\frac{1}{2}M{v}^2{}_s+\frac{1}{2}{I}_s{\omega }^2$$\mathrm{...}(1)$

Condition for rolling without slipping:

  $v_s=r_o\omega$$\mathrm{...}(2)$

Centrifugal power of billiard ball,

  $F_B=m_o\frac{v_s{}^2}{r}$

Calculation:

Let's consider the motion of sphere,

Initially the billiard ball is at rest with translational kinetic energy, so $K_i=\frac{1}{2}{m}_0{v}_0{}^2$.

The initial gravitational potential energy is $U_i=m_{o}g(0)=0$

Final gravitational potential energy is $U_f=m_{o}g(h)$

Conservation of mechanical energy:

  $K_i+U_i=K_f+U_f$

  $\frac{1}{2}{m}_0{v}_0{}^2+(0)=K_f+m_{o}g(h)$

  $K_f=\frac{1}{2}{m}_o{v}_o{}^2-m_{o}gh$$\mathrm{...}(A)$

Let's consider the kinetic energy $\left(K_f\right)$ of billiard ball at the top of the hill,

Kinetic energy is part translational and part rotational. We can use $(2)$ equation to write $\omega$

In terms of $v_s$.

Using $(1)$ expression,

Kinetic energy for an object that undergoes both translation and rotation:

  $K=\frac{1}{2}M{v}^2{}_s+\frac{1}{2}{I}_s{\omega }^2$$\mathrm{...}(1)$

  $K_f=\frac{1}{2}{m}_o{v}^2{}_s+\frac{1}{2}{I}_s{\omega }^2$

Condition for rolling without slipping:

  $v_s=r_o\omega$$\mathrm{...}(2)$

  $\omega =\frac{v_s}{r_o}$

Substitute into kinetic energy equation:

  $K_f=\frac{1}{2}{m}_o{v}^2{}_c+\frac{1}{2}{I}_c{\omega }^2$

  $K_f=\frac{1}{2}{m}_o{v}^2{}_s+\frac{1}{2}{I}_s{\left(\frac{v_s}{r_o}\right)}^2$

  $K_f=\frac{1}{2}\left(m_o+\frac{I_s}{r_0{}^2}\right)v_s{}^2$

From the general knowledge we know that moment of inertia of a sphere is $\frac{2}{5}{m}_0{r}_o{}^2$.

So, let's substitute the $I_s$ value in to the equation,

  $K_f=\frac{1}{2}\left(m_o+\frac{I_s}{r_0{}^2}\right)v_s{}^2$

  $K_f=\frac{1}{2}\left(m_o+\frac{\left(\frac{2}{5}{m}_0{r}_o{}^2\right)}{r_o{}^2}\right)v_s{}^2$

  $K_f=\frac{1}{2}\left(m_o+\frac{2}{5}{m}_0\right)v_s{}^2$

  $K_f=\frac{1}{2}\left(\frac{7}{5}{m}_0\right)v_s{}^2$

  $K_f=\frac{7}{10}{m}_o{v}_s{}^2$$\mathrm{...}(B)$

Since $(A),(B)$ equations are equal,

  $(A)=(B)$

  $\frac{1}{2}{m}_0{v}_0{}^2-m_{o}g(h)=\frac{7}{10}{m}_o{v}_s{}^2$

  $\frac{1}{2}{v}_0{}^2-g(h)=\frac{7}{10}{v}_s{}^2$

  $\frac{10}{7}\left(\frac{1}{2}{v}_0{}^2-g(h)\right)=v_s{}^2$$\mathrm{...}(3)$

Let's consider the billiard ball at the top of the hill,

Centrifugal power of billiard ball,

  $F_B=m_o\frac{v_s{}^2}{r}$

Substituting $v_s{}^2$ value to above equation,

  $F_B=m_o\frac{v_s{}^2}{r}$

  $F_B=m_o\frac{\frac{10}{7}\left(\frac{1}{2}{v}_0{}^2-g(h)\right)}{r}$

  $F_B=m_o\frac{10\left(v_0{}^2-2g(h)\right)}{14r}$

Let's substitute the values,

  $F_B=m_o\frac{10\left(v_0{}^2-2g(h)\right)}{14r}$

  $F_B=m_o\frac{10\left({(2.00m{s}^{-1})}^2-2*10m{s}^{-2}*(-10*{10}^{-2}m)\right)}{14r}$

  $F_B=m_o\frac{10\left((4.00m^2{s}^{-2}+2.00m^2{s}^{-2}\right)}{14r}$

  $F_B=\frac{160*{10}^{-3}kg*10\left(6.00m^2{s}^{-2}\right)}{14*60*{10}^{-2}m}$

  $F_B=\frac{16}{14}N$

  $F_B=1.143N$

Conclusion:

Thus, normal force acting on the billiard ball at point $B$ is $1.143N$.