A particle of mass m is kept on a fixed, smooth sphere of...

A particle of mass $m$ is kept on a fixed, smooth sphere of radius $R$ at a position, where the radius through the particle makes an angle of $30^{\circ}$ with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release? (b) Find the distance travelled by the particle before it leaves contact with the sphere.

Key Concepts

Energy Conservation in Gravitational Systems

The principle of conservation of energy relates the potential energy lost by the particle as it descends to the kinetic energy gained. This is instrumental in relating the particle’s speed at various positions on the sphere to its height, thereby linking energy considerations with dynamics on the curved surface.

Centripetal Force in Curved Motion

For an object moving along a curved path, such as on a sphere, a centripetal force is necessary to maintain the circular motion. This force is provided by the net radial component of the gravitational force and the normal force, and it must equal the mass times the square of the velocity divided by the radius of curvature.

Decomposition of Forces

In problems involving motion on a curved surface, the gravitational force is resolved into components perpendicular (normal) and parallel (tangential) to the surface. The normal component affects the contact force provided by the surface, while the tangential component governs the acceleration along the surface.

Normal Force and Contact Condition

The normal force is the force exerted by the surface that acts perpendicular to it, countering the perpendicular component of gravity. In such problems, the particle loses contact with the surface when this normal force becomes zero, making it critical to analyze the balance of forces that provide the required centripetal acceleration.

Transcript

00:01 In this question, we have a smooth sphere of radius r.

00:04 A particle with mass m is held at an angle of 30 degrees to the vertical.

00:09 It is then released.

00:11 The first part of this question wants us to find out the force exerted on the particle by the sphere just after it's released.

00:18 So we know at this point, this must be equal to the normal contact force on the sphere.

00:24 As a result, we must resolve the weight of the sphere in the direction.

00:30 Direction of the radius to find out the normal contact force.

00:34 So what we know is that this angle in here is 30 degrees.

00:39 This is the radius and as the normal contact force acts vertically downwards, this angle in here must also equal 30 as it's parallel to the vertical, so due to alternate angles is equal to 30 degrees.

00:51 So as a result, for the first part of this question, we can say that the force acting on the small board due to the sphere is equal to the radial component of the weight of the small particle, so is equal to m g, cosine, cosine 30, which is equal to root 3 over 2, mg.

01:24 So that is our first answer.

01:27 The next part wants us to find out at which angle the ball leaves contact with the large sphere.

01:33 So this is the point that the ball is no longer in contact with the surface, and therefore the normal contact force must equal zero.

01:44 As we know that the normal contact force is equal to zero, this means that the resultant radial component of the weight of the small particle must equal the centripetal force to keep it moving in a circle.

01:59 So we can equate these two expressions and say that centripetal force, so mv squared over r, is equal to m g cos theta.

02:13 We can simplify this expression by canceling out these ms, and what we can do is take this r up to this side to obtain an expression for v squared.

02:26 So this gives us that v squared is equal to gr cos theta.

02:37 So this is our first expression.

02:42 The next thing we can say is that we know energy must always be conserved.

02:47 Since the ball starts at rest on the large sphere, it has an initial kinetic energy equal to zero...

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