Pebble dropped on rotating wheel, starts to slide after rotation

[zibs.shirsh]

Homework Statement

A wheel of radius R=50cm rolls along the ground with velocity V=2m/s. A pebble released on top of the wheel so that it is instantaneously at rest on the wheel. The co-efficient of friction between wheel and pebble is μ=1. The pebble starts to slide down when it has rotated through angle θ with the vertical axis.

Calculate θ.

Homework Equations

Absolutely No idea

The Attempt at a Solution

Couldn't attempt

 

[Bryson]

Try drawing a force diagram. This always helps. I am sure you can think of some relevant equations.

 

[haruspex]

I assume it means at rest relative to the point on the wheel where it is placed.
When the wheel has moved through an angle theta (with the pebble not having slipped yet):
- what point is the pebble rotating about?
- how fast is it moving?
- what force is required to keep it rotating about that point?
- what forces act on the pebble, and in what directions?

 

[haruspex]

I assume it means at rest relative to the point on the wheel where it is placed.
When the wheel has moved through an angle theta (with the pebble not having slipped yet):
- what point is the pebble rotating about?
- how fast is it moving?
- what force is required to keep it rotating about that point?
- what forces act on the pebble, and in what directions?

 

[Nickie]

it as no equations were provided.

I would first gather more information about the problem and the variables involved. From the given information, we know that the wheel has a radius of 50cm and is moving with a velocity of 2m/s. The pebble is released on top of the wheel and has a coefficient of friction of 1 with the wheel. We are asked to calculate the angle θ at which the pebble starts to slide down the wheel.

To solve this problem, we need to use the equations of rotational motion and the concept of centripetal force. The equations of rotational motion are:

ω = ω0 + αt (1)
θ = ω0t + 1/2αt^2 (2)
ω^2 = ω0^2 + 2αθ (3)

where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, t is time, and θ is the angle rotated.

We also know that the centripetal force, Fc, is given by:

Fc = mv^2/r (4)

where m is the mass of the pebble, v is the velocity of the wheel, and r is the radius of the wheel.

To solve for θ, we can use equations (1) and (4). We know that at the point where the pebble starts to slide, the centripetal force is equal to the frictional force, which is given by:

Ff = μN (5)

where μ is the coefficient of friction and N is the normal force.

Substituting equations (4) and (5) into (1), we get:

μmg = mv^2/r (6)

Solving for θ in equation (2), we get:

θ = ω0t + 1/2αt^2 (7)

Substituting equations (6) and (7) into (3), we get:

ω^2 = ω0^2 + 2αθ (8)

Solving for ω, we get:

ω = √(ω0^2 + 2αθ) (9)

Substituting equation (9) into (6), we get:

μmg = m(ω0^2 + 2αθ)r (10)

Solving for θ, we get:

θ = (