Question

A solid sphere of mass 0.1 kg and radius 2.5 cm rolls without sliding with a uniform velocity of 0.1 ms-1 along a straight line on a smooth horizontal table. Find its total energy.

Answer 1

Answer:

7×10^-4

Explanation:

Total energy will be equal to the sum of translational and rotational kinetic energies.

Total KE= Translational KE + Rotational KE

$translational \: ke \: = \frac{1}{2} m {v}^{2} \\ rotational \: ke = \frac{1}{2} i {w}^{2}$

where,

• m=mass
• v=translational velocity
• i=moment of inertia
• w=Angular velocity.

$i \: of \: solid \: sphere = \frac{2}{5}m {r}^{2} = \frac{2}{5} \times 0.1 \times {(2.5 \times {10}^{ - 2)} }^{2} = 2.5 \times {10}^{ - 5} \\ w = \frac{v}{r} = \frac{0.1}{(2.5 \times {10}^{ - 2} )} = 4 \\ \\ \\ \frac{1}{2} m {v}^{2} = \frac{1}{2} \times 0.1 \times {0.1}^{2} \\ = 5 \times {10}^{ - 4} \\ \\ \\ \frac{1}{2} i {w}^{2} = \frac{1}{2} 2.5 \times {10}^{ - 5} \times {4}^{2} \\ = 2 \times {10}^{ - 4}$

therefore, total energy is given by,

translational KE + rotational KE

i.e,

$5 \times {10}^{ - 4} + 2 \times {10}^{ - 4} \\ = 7 \times {10}^{ - 4}$