Chapter 8, Problem 67QAP

To determine

Cylinder's speed at the bottom of the ramp.

Answer to Problem 67QAP

  $v_c=3.36m{s}^{-1}$

Explanation of Solution

Given info:

Uniform solid cylinder of,

  Radius $=r_0=5cm$

  Mass $=\theta ={25}^{\circ }$

  $=m_0=3.00kg$

An inclined plane,

  Length $=l_0=2.00m$

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

Cylinder rolls without slipping down the ramp.

Formula used:

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

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

Let's name the linear speed of cylinder at the bottom of the ramp as $v_c$

Let's name the moment of inertia of cylinder as $I_c$.

  $g=10m{s}^{-2}$.

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{}_c+\frac{1}{2}{I}_c{\omega }^2$$\mathrm{...}(1)$

Condition for rolling without slipping:

  $v_c=r_o\omega$$\mathrm{...}(2)$

Calculation:

Let's consider the motion of cylinder,

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

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:

  $K_i+U_i=K_f+U_f$

  $(0)+m_{o}g{y}_i=K_f+m_{o}g{y}_f$

  $K_f=m_{o}g(y_i-y_f)$

But, according to the data given,

  $(y_i-y_f)=h$

So,

  $K_f=m_{o}gh$$\mathrm{...}(A)$

Let's consider the kinetic energy $\left(K_f\right)$ of cylinder 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 $v_c$.

Using $(1)$ expression,

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

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

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

Condition for rolling without slipping:

  $v_c=r_o\omega$$\mathrm{...}(2)$

  $\omega =\frac{v_c}{r_o}$

Substitute into kinetic energy equation:

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

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

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

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

From the general knowledge we know that moment of inertia of a cylinder is $I_c=\frac{1}{2}{m}_0{r}_o{}^2$.

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

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

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

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

  $K_f=\frac{1}{2}\left(\frac{3}{2}{m}_0\right)v_c{}^2$

  $K_f=\frac{3}{4}{m}_o{v}_c{}^2$$\mathrm{...}(B)$

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

  $(A)=(B)$

  $m_{o}gh=\frac{3}{4}{m}_o{v}_c{}^2$

  $gh=\frac{3}{4}{v}_c{}^2$

  $v_c{}^2=\frac{4gh}{3}$

  $v_c=\sqrt{\frac{4gh}{3}}$

Let's substitute the values,

  $v_c=\sqrt{\frac{4*10m{s}^{-2}*2m\sin {25}^{\circ }}{3}}$

  $v_c=\sqrt{11.27}$

  $v_c\approx 3.36m{s}^{-1}$

  $v_c=3.36m{s}^{-1}$

Conclusion:

Thus, cylinder's speed at the bottom of the ramp is $3.36m{s}^{-1}$.