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.)
h =
Ac at top of loop = g = v^2/R
so
v^2 at top of loop must be g R
Height difference = (h-2R)
so total Ke at top of loop = m g (h-2R)
but
total Ke = (1/2)m v^2 + (1/2)I w^2
but v = w r so w = v/r
total Ke = (1/2) [m + I/r^2]v^2
but
v^2 must be gR
so
total Ke = (1/2)[m + I/r^2]gR
but we know total Ke is loss of Pe = m g (h-2R)
so
(h-2R) = (1/2)[ 1 + I/mr^2] R
h = 2R + (1/2)[ 1 + I/mr^2] R
I is (2/5) m r^2
so
h = 2R + (1/2) [ 7/5 ]R
h = 2.7 R
h = your confidence in my ability to solve physics problems without falling flat on my face * (m*g) / (R + r)
To determine the height from which the ball should be launched in order to make it through the loop without falling off the track, we can use the law of conservation of energy.
The initial potential energy of the ball when it is launched from a height h is given by mgh, where m is the mass of the ball and g is the acceleration due to gravity.
The final kinetic energy of the ball when it is at the top of the loop is given by 1/2 mv^2, where v is the velocity of the ball.
Since the ball is rolling without slipping, the velocity can be related to the radius of the loop and the angular velocity of the ball. The angular velocity ω is related to the velocity v and the radius r by v = ωr.
Using the conservation of energy, we set the initial potential energy equal to the final kinetic energy:
mgh = 1/2 mv^2
Substituting v = ωr, we have:
mgh = 1/2 m(ωr)^2
Simplifying and canceling out the mass m:
gh = 1/2 (ωr)^2
Next, we can relate the angular velocity ω to the radius R and the height h. At the top of the loop, the ball is in pure rolling motion, so the centripetal acceleration a = ω^2R.
Using the relation between acceleration, velocity, and radius:
a = ω^2R = g
Substituting this into the previous equation:
gh = 1/2 gR^2
Solving for h:
h = 1/2 R^2
Therefore, the ball should be launched from a height h equal to 1/2 the radius squared in order to make it through the loop without falling off the track.
To determine the height from which the ball should be launched in order to make it through the loop without falling off the track, we can equate the gravitational potential energy at the top of the loop to the sum of the kinetic energy and the gravitational potential energy at the initial position.
Let's break down the problem and explain the steps to find the solution:
1. At the top of the loop, the ball is at its highest point. Here, it has gravitational potential energy and no kinetic energy.
2. At the initial position (when launched), the ball has a certain height h. Here, it has gravitational potential energy and initial kinetic energy due to its translational motion.
3. The condition for the ball to remain on the track and complete the loop is that the centripetal force provided by gravity must not exceed the maximum static friction.
4. The centripetal force at the top of the loop is provided by the normal force (N) subtended by the loop radius (R) and the gravitational force (mg) of the ball.
5. The normal force can be calculated as N = mg + mv²/R, where v is the velocity of the ball at the top of the loop. This equation uses the centripetal acceleration, a = v²/R, and the relation between the force and acceleration, F = ma.
6. At the top of the loop, the normal force minus the weight (mg) supplies the centripetal force, resulting in the net inward force for the ball.
7. This net inward force acts as the maximum static friction, Ff (f = μsN), where μs is the coefficient of static friction between the ball and the track.
8. Since we want the ball to complete the loop without falling off, the maximum static friction must be sufficient to provide the inward force.
9. The highest point (top of the loop) is where the static friction is maximum, and thus we consider this condition.
10. Equating the gravitational potential energy at the top of the loop to the sum of the kinetic energy and gravitational potential energy at the initial position, we have:
mgh = 1/2mv² + mgh
Simplifying the equation, we get:
gh = 1/2v²
11. Rearranging the equation to solve for h, we have:
h = (1/2v²)/g
Now, using the given variables and the equations derived, we can calculate the value of h.