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प्रश्न
A hollow sphere and a solid sphere having same mss and same radii are rolled down a rough inclined plane.
पर्याय
The hollow sphere reaches the bottom first.
The solid sphere reaches the bottom with greater speed
The solid sphere reaches the bottom with greater kinetic energy.
The two spheres will reach the bottom with same linear momentum.
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उत्तर
The solid sphere reaches the bottom with greater speed.
Acceleration of a sphere on the incline plane is given by
\[a = \frac{g\sin\theta}{1 + \frac{I_{COM}}{m r^2}}\]
\[ I_{COM}\] for a solid sphere \[= \frac{2}{5}m r^2 \]
\[\text{So, }a = \frac{g\sin\theta}{1 + \frac{2m r^2}{5m r^2}} = \frac{5}{7}g\sin\theta\]
\[I_{COM}\] for a hollow sphere \[= \frac{2}{3}m r^2 \]
\[\text{So, }a' = \frac{g\sin\theta}{1 + \frac{2m r^2}{3m r^2}} = \frac{3}{5}g\sin\theta\]
The acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.
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संबंधित प्रश्न
Derive an expression for kinetic energy, when a rigid body is rolling on a horizontal surface without slipping. Hence find kinetic energy for a solid sphere.
Read each statement below carefully, and state, with reasons, if it is true or false;
A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion
A sphere cannot roll on
In rear-wheel drive cars, the engine rotates the rear wheels and the front wheels rotate only because the car moves. If such a car accelerates on a horizontal road the friction
(a) on the rear wheels is in the forward direction
(b) on the front wheels is in the backward direction
(c) on the rear wheels has larger magnitude than the friction on the front wheels
(d) on the car is in the backward direction.
A sphere can roll on a surface inclined at an angle θ if the friction coefficient is more than \[\frac{2}{7}g \tan\theta.\] Suppose the friction coefficient is \[\frac{1}{7}g\ tan\theta.\] If a sphere is released from rest on the incline, _____________ .
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?
Answer in Brief:
A rigid object is rolling down an inclined plane derive the expression for the acceleration along the track and the speed after falling through a certain vertical distance.
The speed of a solid sphere after rolling down from rest without sliding on an inclined plane of vertical height h is, ______
What is the difference between sliding and slipping?
A solid sphere rolls down from top of inclined plane, 7m high, without slipping. Its linear speed at the foot of plane is ______. (g = 10 m/s2)
A man is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity `omega`. The tension in the strings is F. If the length of string and angular velocity are doubled, the tension in string is now ____________.
A solid sphere is rolling on a frictionless surface with translational velocity 'V'. It climbs the inclined plane from 'A' to 'B' and then moves away from Bon the smooth horizontal surface. The value of 'V' should be ______.
The angular velocity of minute hand of a clock in degree per second is ______.
A 1000 kg car has four 10 kg wheels. When the car is moving, fraction of total K.E. of the car due to rotation of the wheels about their axles is nearly (Assume wheels be uniform disc)
A solid spherical ball rolls on an inclined plane without slipping. The ratio of rotational energy and total energy is ______.
A uniform disc of radius R, is resting on a table on its rim.The coefficient of friction between disc and table is µ (Figure). Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?
The least coefficient of friction for an inclined plane inclined at angle α with horizontal in order that a solid cylinder will roll down without slipping is ______.
The kinetic energy and angular momentum of a body rotating with constant angular velocity are E and L. What does `(2E)/L` represent?
