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Question

Physics Question on System of Particles & Rotational Motion

A ball of radius 11 cm and mass 8 kg rolls from rest down a ramp of length 2 m. The ramp is inclined at 35? to the horizontal. When the ball reaches the bottom, its velocity is (Take, sin 35? = 0.57 and cos 35? = 0.81)

A

2ms1{ 2 \, m \, s^{-1}}

B

5ms1{ 5 \, m \, s^{-1}}

C

4ms1{ 4 \, m \, s^{-1}}

D

6ms1{ 6 \, m \, s^{-1}}

Answer

4ms1{ 4 \, m \, s^{-1}}

Explanation

Solution

Kinetic energy K=12mv2+12Iω2K = \frac{1}{2} mv^2 + \frac{1}{2} I \, \omega^2
K=12mv2+12×(25mr2)ω2K = \frac{1}{2} mv^{2}+\frac{1}{2} \times\left(\frac{2}{5} mr^{2}\right)\omega^{2}
=12mv2+15mv2=710mv2=\frac{1}{2} mv^{2} + \frac{1}{5} mv^{2} = \frac{7}{10} mv^{2} (vrω)(\because \: v - r \omega)
According to conservation of energy, we get
710mv2=mgh\frac{7}{10} mv^2 = mgh

v=107gh=107glsinθv = \sqrt{\frac{10}{7}gh} =\sqrt{\frac{10}{7}gl \sin \theta}
=107×9.8×2sin35=4ms1=\sqrt{\frac{10}{7} \times9.8\times2 \sin 35^{\circ}} = 4 \, { m \, s^{-1}}