A marble of mass m is placed on top of a bowling ball of radius r and allowed to roll from rest. derive the angle from centre of circle at which the marble loses contact with bowling ball. assume there is no friction between marble and bowling ball.
If there were no friction, the marble would slide, not roll. It is not clear which way they want this problem to be done.
In either case, a formula for the velocity of the marble vs. angle from the top can be derived. The rolling and sliding (no friction) cases will be different.
The marble will remain in contact with the bowling ball until
M g cos theta = M V^2/R
cos theta = V^2/(g R)
Knowing V(theta), you can solve for theta.
Although friction is needed to make the ball roll, it does not remove kinetic energy, because there is no "slipping" motion at the point of contact.
To find the angle at which the marble loses contact with the bowling ball, we can analyze the forces acting on the marble.
Since there is no friction between the marble and the bowling ball, the only force acting on the marble is the gravitational force. This force can be split into two components: one acting radially inward towards the center of the ball, and another acting tangentially, causing the marble to roll.
Let's consider a free body diagram of the marble at a point where it's about to lose contact with the bowling ball. At this point, the normal force exerted by the bowling ball on the marble becomes zero, so only the radial component of the gravitational force is acting.
The radial component of the gravitational force can be found using trigonometry. The angle between the radial direction and the vertical direction is equal to the angle from the center of the circle at which the marble loses contact.
Applying trigonometry, we know that the radial component of the gravitational force is equal to the gravitational force multiplied by the cosine of the angle.
Radial force = mg * cos(θ)
where m is the mass of the marble, g is the acceleration due to gravity, and θ is the angle from the center of the circle.
At the point where the marble loses contact, the radial force is equal to the centripetal force required to keep the marble moving in a circle. The centripetal force can be expressed as:
Centripetal force = m * v^2 / r
where v is the velocity of the marble and r is the radius of the bowling ball.
Equating the radial force to the centripetal force, we have:
mg * cos(θ) = m * v^2 / r
Simplifying and canceling out the mass terms, we get:
g * cos(θ) = v^2 / r
Now, we can use the fact that the marble is rolling without slipping to relate the velocity to the radius. When a rolling object does not slip, the velocity of the object is related to the angular velocity (ω) and the radius (r) by the equation:
v = ω * r
Let's substitute this expression for v into our equation:
g * cos(θ) = (ω * r)^2 / r
Simplifying further:
g * cos(θ) = ω^2 * r
We know that the angular velocity can be expressed as:
ω = v / r
Substituting this into our equation, we get:
g * cos(θ) = (v / r)^2 * r
Simplifying again:
g * cos(θ) = v^2 / r
This equation is identical to the equation we obtained earlier for the centripetal force. Therefore, we can conclude that the angle from the center of the circle at which the marble loses contact with the bowling ball is the same angle at which the centripetal force is equal to the radial component of the gravitational force.
So, the angle can be found as:
cos(θ) = v^2 / (gr)
θ = acos(v^2 / (gr))
where a is the inverse cosine function.
Keep in mind that this derivation assumes that the marble and bowling ball are idealized as point masses and neglects any other forces, such as air resistance.