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प्रश्न
A solid sphere of mass 0⋅50 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?
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उत्तर
Let α be the angular acceleration produced in the sphere.
Rotational equation of motion,
\[F \times R - f_r \times R = I\alpha\]
\[\Rightarrow F = \frac{2}{5}mR\alpha + \mu mg........(1)\]
Translational equation of motion,
\[F = ma - \mu mg\]
\[ \Rightarrow a = \frac{\left( F + \mu mg \right)}{m}\]
For pure rolling, we have
\[\alpha = \frac{a}{R}\]
\[\Rightarrow \alpha = \frac{\left( F + \mu mg \right)}{mR}\]
Putting the value of \[\alpha\] in equation (1), we get
\[F = \frac{2}{5}\frac{mR\left( F + \mu mg \right)}{mR} + \mu mg\]
\[ \Rightarrow F = \frac{2}{5}\left( F + \mu mg \right) \mu mg\]
\[ \Rightarrow F = \frac{2}{5}F + \left( \frac{2}{5} \times \frac{2}{7} \times 0 . 5 \times 10 \right) + \left( \frac{2}{7} \times 0 . 5 \times 10 \right)\]
\[ \Rightarrow \frac{3F}{5} = \frac{4}{7} + \frac{10}{7} = 2\]
\[ \Rightarrow F = \frac{5 \times 2}{3} = \frac{10}{3} = 3 . 3 N\]
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संबंधित प्रश्न
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? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?
Read each statement below carefully, and state, with reasons, if it is true or false;
The instantaneous speed of the point of contact during rolling is zero.
Read each statement below carefully, and state, with reasons, if it is true or false;
For perfect rolling motion, work done against friction is zero.
If a rigid body of radius ‘R’ starts from rest and rolls down an inclined plane of inclination
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Answer in Brief:
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