Question

$1)$A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination,
$A)$ Will it reach the bottom with the same speed in each case?

$2)$ A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination,
$B)$ Will it take longer roll down one plane than the other?

$3)$ A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination,
$C)$ If so, which one and why?

Answer 1

$A)$ Step 1: Conservation of energy

From figure,

${\theta }_1>{\theta }_2$ i.e.,$\sin {\theta }_1>\sin {\theta }_2$ (or) $\frac{\sin {\theta }_1}{\sin {\theta }_2}>1$ ………$\left(1\right)$

Where, $h-$ is height of each inclined plane $=l_{1}sin{\theta }_1=l_2\sin {\theta }_2$

Yes, at the top of the planes, the sphere only has $P.E=mgh$

Where, $m$ - is mass of the sphere

When the sphere rolls down from the top to the bottom, $P.E$ is converted into $K.E.$

$\therefore$ If $v_1$ and $v_2$ be its linear speed at the bottom of the planes,

i.e., $\left(1\right)$ and $\left(2\right)$ respectively then,


$mgh=\frac{1}{2}m{v}_1^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}_2^2+\frac{1}{2}I{\omega }^2$

$mgh=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{k}^2\frac{v_1^2}{R^2}=\frac{1}{2}m{v}_2^2+\frac{1}{2}m{k}^2\frac{v_2^2}{R^2}$
$(\therefore v=R\omega \text{and}I=m{k}^2)$

$2gh=v_1^2(1+\frac{k^2}{R^2})$ 𝑎𝑛𝑑 $2gh=v_2^2(1+\frac{k^2}{R^2})$

Therefore,

$v_1^2=\frac{2gh}{(1+\frac{k^2}{R^2})}$ ..........$\left(2\right)$

$v_2^2=\frac{2gh}{(1+\frac{k^2}{R^2})}$ .........$\left(3\right)$

Where, $I-$ is moment of inertia of the sphere its angular speed and $k-$ is the Radius of gyration.

From eq$\left(2\right)$ and $\left(3\right)$ it is clear that, the sphere reaches the bottom with same speed in each case.

$B)$ Yes, it will take longer time down one plane than the other. it will be longer for the plane having smaller angle of inclination.

$C)$ Step 1: Acceleration of the solid sphere

Let $t_1$ and $t_2$ be the time taken by the sphere in rolling on plane $\left(1\right)$ and $\left(2\right)$ respectively. Acceleration of the solid sphere on an inclined plane is given by,

$a=\frac{g\sin \theta }{(1+\frac{k^2}{R^2})}$

Now for solid sphere,
$a=1+\frac{k^2}{R^2}=1+\frac{\frac{2}{5}{R}^2}{R^2}=\frac{7}{5}$

If $a_1$ 𝑎𝑛𝑑 $a_2$ be the accelerations of the sphere on inclined plane $\left(1\right)$ and $\left(2\right)$ respectively, then

$a_1=\frac{g\sin {\theta }_1}{\frac{7}{5}}=\frac{5}{7}g\sin {\theta }_1$


Step 1: Acceleration of the solid sphere

Similarly,

$a_2=\frac{g\sin {\theta }_2}{\frac{7}{5}}=\frac{5}{7}g\sin {\theta }_2$

Step 2: Distance travelled by the sphere

From Kinematic equations of motion,

$S=ut+\frac{1}{2}a{t}^2$

We get, $t_1^2=\frac{2l_1}{a_1}=\frac{\frac{2h}{\sin {\theta }_1}}{\frac{5}{7}g\sin {\theta }_1}=\frac{14h}{5g(\sin {\theta }_1{)}^2}$ ..$\left(4\right)$

and $t_2^2=\frac{2l_2}{a_2}=\frac{\frac{2h}{\sin {\theta }_2}}{\frac{5}{7}g\sin {\theta }_2}=\frac{14h}{5g(\sin {\theta }_2{)}^2}$ ..$\left(5\right)$


Dividing eq $\left(4\right)$ and $\left(5\right)$ we get,

$\frac{t_1^2}{t_2^2}=\frac{(\sin {\theta }_2{)}^2}{(\sin {\theta }_1{)}^2}\therefore \frac{t_1}{t_2}=\frac{\sin {\theta }_2}{\sin {\theta }_1}$ .......$\left(6\right)$

Now, eq $\left(1\right)$

$\frac{t_1}{t_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}>1\left(or\right)\frac{\sin {\theta }_2}{\sin {\theta }_1}<1$ $\therefore {\left(\frac{\sin {\theta }_2}{\sin {\theta }_1}\right)}^2<1$ .......$\left(7\right)$

$\therefore$ From eq $\left(6\right)$ and $\left(7\right)$ we get,

$\frac{t_1}{t_2}<1\left(or\right)t_1<1$

It is due to the reason that $a\propto sin\theta$ and $t$ is inversely proportional to $‘a^{\prime }$ or $sin\theta$