Chapter 8, Problem 68QAP

To determine

Sphere's speed at the bottom of the ramp.

Answer to Problem 68QAP

  $v_s=3.86m{s}^{-1}$

Explanation of Solution

Given info:

Uniform solid sphere of,

  Radius $=r_0=5cm$

  Mass $=m_0=3.00kg$

An inclined plane,

  Length $=l_0=2.00m$

  Angle which tilted with horizontal plane $=\theta ={25}^{\circ }$

Sphere rolls without slipping down the ramp.

Translational speed of sphere at the top of plane $=v_o=2.00m{s}^{-1}$

Formula used:

Let's name the vertical height of the plane as $h$.

Let's name the angular velocity of sphere at the bottom of the ramp as $\omega$.

Let's name the linear speed of sphere at the bottom of the ramp as $v_s$

Let's name the moment of inertia of sphere 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)$

Calculation:

Let's consider the motion of sphere,

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

The initial gravitational potential energy is $U_i=m_{o}g{y}_i$

Final gravitational potential energy is $U_f=m_{o}g{y}_f$

Conservation of mechanical energy:

  $K_i+U_i=K_f+U_f$

  $\frac{1}{2}{m}_0{v}_0{}^2+m_{o}g{y}_i=K_f+m_{o}g{y}_f$

  $K_f=m_{o}g(y_i-y_f)+\frac{1}{2}{m}_0{v}_0{}^2$

But, according to the data given,

  $(y_i-y_f)=h$

So,

  $K_f=m_{o}gh+\frac{1}{2}{m}_0{v}_0{}^2$$\mathrm{...}(A)$

Let's consider the kinetic energy $\left(K_f\right)$ of sphere at the bottom of the ramp,

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,

  $(A)=(B)$

  $m_{o}gh+\frac{1}{2}{m}_0{v}_0{}^2=\frac{7}{10}{m}_o{v}_s{}^2$

  $gh+\frac{1}{2}{v}_0{}^2=\frac{7}{10}{v}_s{}^2$

  $v_s{}^2=\frac{10}{7}\left(gh+\frac{1}{2}{v}_0{}^2\right)$

  $v_s=\sqrt{\frac{10}{7}\left(gh+\frac{1}{2}{v}_0{}^2\right)}$

Let's substitute the values,

  $v_s=\sqrt{\frac{10}{7}\left(10m{s}^{-2}*2.00m\sin {25}^{\circ }+\frac{1}{2}{(2.00m{s}^{-1})}^2\right)}$

  $v_s=\sqrt{\frac{10}{7}\left(8.452m^2{s}^{-2}+2.00m^2{s}^{-2}\right)}$

  $v_s=\sqrt{\frac{10}{7}\left(10.452m^2{s}^{-2}\right)}$

  $v_s=\sqrt{\left(14.931m^2{s}^{-2}\right)}$

  $v_s=3.86m{s}^{-1}$

Conclusion:

Thus, sphere's speed at the bottom of the ramp is $3.86m{s}^{-1}$.