Question: In Figure, a solid cylinder of radius 10 and mass 12 kgstarts from rest and rolls without slipping a distance L =6.0 mdown a roof that is inclined at the angle $\theta ={30}^0$. (a) What is the angular speed of the cylinder about its centre as it leaves the roof? (b) The roof’s edge is at height H = 5.0mHow far horizontally from the roof’s edge does the cylinder hit the level ground?

The radius of the cylinder,$r=10\text{cm}=0.1\text{m}$

The mass of the cylinder, m = 12 kg

$\text{Slippingdistance}=6.0\text{m}$

Inclination of roof,$\theta ={30}^0$

$\text{Heightofrooffromground}=5.0\text{m}$

As the ball is rolling from a height to the ground, use the conservation of energy to calculate the angular speed of the cylinder. From the value of angular speed, calculate the velocity at which it leaves the roof which will be the initial velocity for the next part. The next motion of the cylinder will be projectile motion, so using kinematic equations, find the horizontal distance.

The linear velocity in terms of the angular velocity and radius is given as-

$\mathbf{v}\mathbf{=}\mathbf{r}\mathbf{\times }\mathbf{\omega }$

The kinetic energy is given as-

${\mathbf{K}}_{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Mv}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}$

The second equation of motion is given as-

$\mathbf{y}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{at}}^{\mathbf{2}}$

Here,M is the mass, I is the moment of inertia, y is the vertical displacement, a is the acceleration, t is the time taken to reach the ground and v₀ is the velocity of the cylinder as it left the roof.

Let the ball covers a distance along the roof vertically and L be the length covered at an inclination $\theta$, so-

$$\begin{gathered} \sin \theta =\frac{h}{L} \\ h=L\times \sin \theta \\ h=6.0\text{m}\times \sin 30° \end{gathered}$$

According to the energy conservation law-

$$\begin{gathered} \text{change inpotentialenergy}=\text{change inKineticEnergy} \\ mgh=\frac{1}{2}m{r}^2{\omega }^2+\frac{1}{4}m{r}^2{\omega }^2 \\ {\omega }^2{r}^2=\frac{{\displaystyle 4}}{{\displaystyle 2}}gh \\ {\omega }^2=\frac{{\displaystyle 4gh}}{{\displaystyle 3r^2}} \\ \omega =\sqrt{\frac{{\displaystyle 4gh}}{{\displaystyle 3r^2}}} \end{gathered}$$

For the given values-

role="math" localid="1660987717011" $$\begin{gathered} \omega =\frac{1}{r}\times \sqrt{\frac{4gh}{3}} \\ \omega =\frac{1}{0.1\text{m}}\times \sqrt{\frac{4\times 9.8\times 3\text{m}}{3}} \\ \omega =63\text{rad/s} \end{gathered}$$

$$\begin{gathered} v=r\omega \omega \\ =0.1m\times 63rad/s \\ =6.3m/s \end{gathered}$$

As the plane is inclined at 30, the cylinder will leave the roof at the same angle. The horizontal and vertical component of velocity is given as-

$$\begin{gathered} V_{0x}=v_{0}Cos{30}^0 \\ =6.3m/s\times 0.866 \\ =5.45m/s \\ {V}_{0y}=v_{0}Sin{30}^0 \\ =6.3m/s\times 0.5 \\ =3.10m/s \end{gathered}$$

To find the time required to travel the given height, we can use vertical component of the velocity. So we have,

$$\begin{gathered} \text{y}={\text{v}}_{0\text{y}}\text{t}+\frac{1}{2}{\text{at}}^2 \\ 5.0\text{m}=\left(3.1m/s\right)\text{t}+\frac{1}{2}\times 9.8m/s^2\times {\text{t}}^2 \\ \left(4.9m/s^2\right){\text{t}}^2+\left(3.1m/s\right)\text{t}=5.0\text{m} \\ \left(4.9m/s^2\right){\text{t}}^2+\left(3.1m/s\right)\text{t}-5.0\text{m}=0 \end{gathered}$$

By solving the above quadratic equation, we get the two values of t are,

$\text{t}=0.74\text{and}-1.37\text{s}$

As the time cannot be negative, so we take positive value for t, i.e.,

Since there is no acceleration in the horizontal direction, we have

$$\begin{gathered} \text{Speed}=\frac{\text{distance}}{\text{time}} \\ \text{Distance}=\text{speed}\times \text{time} \\ d=5.45m/s\times 0.74\text{s} \\ d=4.0\text{m} \end{gathered}$$

The horizontal distance covered by the cylinder after falling from roof is 4.0 m .