A solid ball of mass m and radius r rolls without slipping through a loop of radius R, as shown in the figure. From what height h should the ball be launched in order to make it through the loop without falling off the track? (Use any variable or symbol stated above along with the following as necessary: g.)
To determine the height h from which the ball should be launched in order to make it through the loop without falling off the track, we need to consider the energy conservation principle. The ball will start with gravitational potential energy at height h and convert it into kinetic energy as it rolls down the track.
1. First, let's consider the potential energy at height h:
- The potential energy of an object at height h is given by the formula: PE = mgh, where m is the mass of the ball and g is the acceleration due to gravity.
2. Next, we need to calculate the kinetic energy of the ball when it reaches the bottom of the loop:
- The kinetic energy of an object is given by the formula: KE = (1/2)mv^2, where m is the mass of the ball and v is its velocity.
- In this case, since the ball is rolling without slipping, its velocity can be related to its angular velocity.
- The relationship between linear and angular velocity for an object rolling without slipping is: v = ωr, where ω is the angular velocity and r is the radius of the ball.
3. Now, let's find the angular velocity at the bottom of the loop:
- The centripetal acceleration of the ball at the bottom of the loop is provided by the normal force and is given by: ac = v^2 / R.
- Since there is no slipping, the frictional force must provide the centripetal force: f_friction = m * ac.
- The frictional force can also be expressed as: f_friction = μ * m * g, where μ is the coefficient of friction.
- Equating the two expressions for f_friction, we have: m * ac = μ * m * g.
- Rearranging the equation and subsituting v = ωr, we get: ω = √(μ * g * r).
4. Combining the equations for kinetic energy and angular velocity, we have:
KE = (1/2)mv^2 = (1/2)m(ωr)^2 = (1/2)m(μ * g * r)^2.
5. Finally, we can solve for the height h:
At the top of the loop, when the ball just loses contact with the track, the centripetal acceleration becomes zero.
Applying the energy conservation principle between the initial height h and the bottom of the loop, we have: PE = KE.
mgh = (1/2)m(μ * g * r)^2.
Solving for h, we get: h = (1/2)(μ * g * r)^2 / g.
Thus, the height h from which the ball should be launched to make it through the loop without falling off the track is given by h = (1/2)(μ * g * r)^2 / g.