A particle of mass m is kept on the top of a smooth sphere...

A particle of mass $m$ is kept on the top of a smooth sphere of radius $R .$ It is given a sharp impulse which imparts it a horizontal speed $v .$ (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of $v$ for which the particle does not slip on the sphere? (c) Assuming the velocity $v$ to be half the minimum calculated in part (b) find the angle made by the radius

Key Concepts

Normal Reaction Force

The normal reaction force is the force exerted by a surface perpendicular to the contacting object. In problems involving motion on a curved surface, the normal force plays a key role in providing (or reducing) the required centripetal force. Its magnitude depends on the balance between gravitational pull and the inertial effects due to the particle’s curved trajectory.

Energy Conservation

The principle of energy conservation relates changes in potential energy to changes in kinetic energy as an object moves under the influence of gravity. For a particle sliding on a curved surface, the gain in kinetic energy comes from the loss of gravitational potential energy. This relation is instrumental in determining the particle’s speed at different positions along its path.

Impulse

An impulse is a short duration force that changes the momentum of an object instantaneously. In mechanics, when an impulse is applied, the velocity of the particle changes immediately while its position remains effectively unchanged during the impulse. This concept helps explain how systems can acquire a sudden speed or direction without the need for a prolonged application of force.

Centripetal Force and Circular Motion

When a particle moves along a curved path, it experiences a centripetal acceleration directed towards the center of curvature. The required centripetal force, given by mv²/R for motion along a circle of radius R, must be provided by the net force acting along the radius. Understanding the balance between this centripetal requirement and the forces available (like gravity and the normal force) is crucial in analyzing motion on curved surfaces.

Condition for Loss of Contact

A particle moving along a curved surface loses contact when the normal force becomes zero. This happens when the component of gravitational force along the radius becomes insufficient to supply the centripetal force needed for the circular motion. Establishing the condition for zero normal force is key in solving problems involving the departure from a curved surface or ensuring continuous contact.

Transcript

00:01 In this question, we have a smooth sphere with a particle of mass m on the sphere.

00:05 It's given a horizontal impulse of v meters per second.

00:09 The first part of this question wants us to find the normal force between the sphere and the particle just after the impulse.

00:15 So we know that right after the impulse is going to have a centripetal force of mv squared over r, which keeps it moving in a circle.

00:26 This must be equal to the component due to gravity, taking away the normal contact force as this acts in the opposite direction.

00:34 The normal contact force will act outwards.

00:37 So we know that mv squared over r must be equal to m g minus n.

00:44 And as a result, we can rearrange this expression for part a and say that n is equal to m g minus mv squared over r.

01:01 Because we are assuming it is in circular motion and the circular motion is produced by the resultant of the mg component, so the weight component and the normal course.

01:12 The second part of this question wants us to find the minimum value for the velocity in which the particle does not slip.

01:19 So now we have to say that this is at the point that it doesn't slip on the surface of the sphere and it instead flies off the surface, which means that the normal contact force is zero because it's no longer in contact with the surface.

01:31 Thus, we use this expression obtained from part a, and we can say that m, this time v minimum, squared over r, is equal to mg.

01:53 Minus n, however, n is equal to zero.

01:56 So as a result, rearranging this expression, canceling these m's and taking this r up to here, we can obtain that the minimum velocity is equal to the square root of rg.

02:14 So this is our answer for the second part...

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