Question

A solid sphere of mass 50g and diameter 2cm rolls without sliding with a uniform velocity of 5ms^-1 along the straight line on a smooth horizontal table.calculate its kinetic energy.
(Hint: total E k =1/2mv^2+1/2 I w^2

Answer 1

5mv hsgjijij hfsuvubujn chuma kyvjh

Answer 2

Answer:

Explanation:

Given,

• Mass of the ball = m = 50 g = 0.050 kg
• Final velocity of the ball at lowest point of the surface = v = 5 m/s.

Moment of inertia of the rotating ball = $I\ =\ \dfrac{2}{5}mR^2}$

Angular speed of the rotating ball = $w\ =\ \dfrac{v}{R}$

Now the Kinetic energy due to the rotation,

$K.E_1\ =\ \dfrac{1}{2}Iw^2\ =\ \dfrac{1}{2}\times \dfrac{2}{5}mR^2\times \dfrac{v^2}{R^2}\ =\ \dfrac{1}{5}mv^2$

Again, the kinetic energy due to the translational motion,

$K.E_2\ =\ \dfrac{1}{2}mv^2$

Now total kinetic energy at the lowest point of the surface

$K.E_t\ =\ K.E_1\ +\ K.E_2\\\Rightarrow K.E_t\ =\ \dfrac{1}{5}mv^2\ +\ \dfrac{1}{2}mv^2\\\Rightarrow K.E_t\ =\ \dfrac{7mv^2}{10}\\\Rightarorw K.E_t\ =\ \dfrac{7\times 0.05\times 5^2}{10}\\\Rightarrow K.E_t\ =\ 0.875\ J$