Chapter 8, Problem 69QAP

To determine

Minimum height that the marble must start from to make it around the loop.

Answer to Problem 69QAP

  $h=0.54m$

Explanation of Solution

Given info:

Spherical marble of,

  Radius $=r_0=0.500cm$

  Mass $=m_o=50.0g$

Loop-the-loop track,

  Radius $=r_l=20.0cm$

Spherical marble rolls without slipping down the track.

Marble starts from rest and just barely clears the loop to emerge on the other side of the track.

Formula used:

Let's name the minimum height that the marble must start from to make it around the loop

As $h$.

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

Let's name the linear speed of spherical marble at the top of the track as $v_s$

Let's name the moment of inertia of spherical marble as $I_s$.

  $g=10m{s}^{-2}$.

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)$

Equilibrium of vertical forces of spherical marble at the top of the track,

Centrifugal power $=$ weight

  $m_o(\frac{v_s{}^2}{r_l})=m_{o}g$

Calculation:

Let's consider the motion of sphere,

Initially the spherical marble is at rest with zero kinetic energy, so $K_i=0$.

The initial gravitational potential energy is $U_i=m_{o}gh$

Final gravitational potential energy is $U_f=m_{o}g(r_l+r_l)=2m_{o}g{r}_l$

Conservation of mechanical energy:

  $K_i+U_i=K_f+U_f$

  $\left(0\right)+m_{o}gh=K_f+2m_{o}g{r}_l$

  $K_f=m_{o}g(h-2r_l)$$\mathrm{...}(A)$

Let's consider the kinetic energy $\left(K_f\right)$ of spherical marble at the top of the track,

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,

  $(A)=(B)$

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

  $m_{o}gh=\frac{7}{10}{m}_o{v}_s{}^2+2m_{o}g{r}_l$

  $gh=\frac{7}{10}{v}_s{}^2+2g{r}_l$$\mathrm{...}(3)$

Marble starts from rest and just barely clears the loop to emerge on the other side of the track.

So,

Equilibrium of vertical forces of spherical marble at the top of the track,

Centrifugal power $=$ weight

  $m_o(\frac{v_s{}^2}{r_l})=m_{o}g$

  $(\frac{v_s{}^2}{r_l})=g$

  $v_s{}^2=r_{l}g$

Substituting $v_s{}^2$ value to $(3)$ equation,

  $gh=\frac{7}{10}{v}_s{}^2+2g{r}_l$

  $gh=\frac{7}{10}(r_{l}g)+2g{r}_l$

  $h=\frac{7}{10}{r}_l+2r_l$

  $h=\frac{27}{10}{r}_l$

Let's substitute the $r_l$ value,

  $h=\frac{27}{10}(0.20m)$

  $h=0.54m$

Conclusion:

Thus, minimum height that the marble must start from to make it around the loop is $0.54m$.