91影视

From the law of conservation of energy we can write,

$\mathit{K}\mathit{E}\mathbf{=}\mathit{P}\mathit{E}$

Where KE is kinetic energy

PE is potential energy

Here we have given that maximum height of solid uniform ball is$h_0$

Let mass of the solid ballm

Let the angular speed of the ball up the hill be$蝇$

We know that, the rotational kinetic energy of the ball is given by the equation,

$K{E}_1=\frac{1}{2}l{蝇}^2$

And, the translational kinetic energy of the ball is given by the equation,

$K{E}_2=\frac{1}{2}m{v}^2$

So, the total kinetic energy of the ball is

$$\begin{gathered} KE=K{E}_1+K{E}_2 \\ KE=\frac{1}{2}l{蝇}^2+\frac{1}{2}m{v}^2 \end{gathered}$$

Wherel is moment of inertia,$蝇$ is angular speed,$m$ is mass of solid ball andv is velocity of solid ball.

Also, the potential energy of the ball is

$PE=mg{h}_0$

Where is the mass of solid ball,g is gravitational force and$h_0$ ismaximum height of solid uniform ball.

Now, from the work energy theorem,

$\frac{1}{2}l{蝇}^2+\frac{1}{2}m{v}^2=mg{h}_0............\left(1\right)$

Now, we know that moment of inertia of solid sphere is$l=\frac{2}{5}m{r}^2$

Also, we know that the angular speed of the ball will be,

$w=\frac{v}{r}$

Now substitute these values in equation (1)

So, above equation becomes

$$\begin{gathered} \frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(蝇\right)}^2+\frac{1}{2}m{v}^2=mg{h}_0 \\ \frac{1}{2}m{r}^2{蝇}^2+\frac{1}{2}m\left(r{蝇}^2\right)=mg{h}_0 \\ \frac{7}{10}=r^2{\left(\frac{v}{r}\right)}^2=g{h}_0 \\ {h}_0=\frac{7v^{2}10g}{}............\left(2\right) \end{gathered}$$

(a)

From equation (2), we have$h_0=\frac{7v^2}{10g}$

It is clear from the above relation that maximum height of the ball does not depend upon the diameter of the ball.

So, there is no change in the maximum height.

(b)

From equation (2), we have$h_0=\frac{7v^2}{10g}$

Here we can see that maximum height does not depend on mass.

So, if mass is doubled, maximum height remains same.

(c)

From equation (2), we have$h_0=\frac{7v^2}{10g}$

Here we can see that maximum height does not depend on either mass or diameter.

So, if both diameter and mass is doubled, maximum height remains same.

(d)

From equation (2), we have$h_0=\frac{7v^2}{10g}$

It is clear from the above relation that maximum height of the ball does not depend upon the angular velocity of the ball.

So, there is no change in the maximum height if the angular velocity is doubled.