Question:
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 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)
Updated On: Mar 9, 2024
- ${ 2 \, m \, s^{-1}}$
- ${ 5 \, m \, s^{-1}}$
- ${ 4 \, m \, s^{-1}}$
- ${ 6 \, m \, s^{-1}}$
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The Correct Option is C
Solution and Explanation
Kinetic energy $K = \frac{1}{2} mv^2 + \frac{1}{2} I \, \omega^2$
$K = \frac{1}{2} mv^{2}+\frac{1}{2} \times\left(\frac{2}{5} mr^{2}\right)\omega^{2} $
$=\frac{1}{2} mv^{2} + \frac{1}{5} mv^{2} = \frac{7}{10} mv^{2} $ $(\because \: v - r \omega)$
According to conservation of energy, we get
$\frac{7}{10} mv^2 = mgh$
$v = \sqrt{\frac{10}{7}gh} =\sqrt{\frac{10}{7}gl \sin \theta} $
$=\sqrt{\frac{10}{7} \times9.8\times2 \sin 35^{\circ}} = 4 \, { m \, s^{-1}}$
$K = \frac{1}{2} mv^{2}+\frac{1}{2} \times\left(\frac{2}{5} mr^{2}\right)\omega^{2} $
$=\frac{1}{2} mv^{2} + \frac{1}{5} mv^{2} = \frac{7}{10} mv^{2} $ $(\because \: v - r \omega)$
According to conservation of energy, we get
$\frac{7}{10} mv^2 = mgh$
$v = \sqrt{\frac{10}{7}gh} =\sqrt{\frac{10}{7}gl \sin \theta} $
$=\sqrt{\frac{10}{7} \times9.8\times2 \sin 35^{\circ}} = 4 \, { m \, s^{-1}}$
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Concepts Used:
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