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 from which the ball should be launched in order to make it through the loop without falling off the track, we can use the conservation of energy.
Let's break down the problem into two parts: the initial position before the ball is launched and the highest position when it reaches the top of the loop.
1. Initial Position:
At the initial point, the ball possesses potential energy (mgh) due to its height from the ground. It does not have any kinetic energy since it is at rest.
2. Highest Position:
When the ball reaches the highest point of the loop, it is momentarily stationary. At this point, all of the ball's initial potential energy is converted into kinetic energy.
To calculate the height (h) from which the ball should be launched, we need to equate the potential energy at the initial position with the kinetic energy at the highest position.
Potential Energy at the initial position (mgh) = Kinetic Energy at the highest point (0.5mv^2 + 0.5Iω^2), where v is the linear velocity of the ball, and ω is the angular velocity of the ball.
For a ball rolling without slipping, we have a relationship between linear and angular velocity:
v = ωr, where r is the radius of the ball.
Also, the moment of inertia (I) for a solid sphere rotating about its diameter is given by:
I = 0.4mr^2
Substituting the above relationships into the equation:
mgh = 0.5mv^2 + 0.5Iω^2
mgh = 0.5m(v^2 + (0.4r^2)ω^2)
mgh = 0.5m(v^2 + (0.4r^2)(v^2/r^2))
mgh = 0.5mv^2 + 0.2mv^2
mgh = 0.7mv^2
Canceling out the mass (m) on both sides of the equation:
gh = 0.7v^2
Now, we need to find the relationship between velocity (v) and radius (R) during a loop. When the ball is at the top of the loop, the net force on it is the centripetal force:
mg - N = mv^2/R, where N is the normal force exerted by the track.
The normal force (N) can be calculated using:
N = mg - mv^2/R
Substitute this value of N into the earlier equation:
mg - (mg - mv^2/R) = mv^2/R
Simplifying the equation:
mv^2/R = mv^2/R
v^2 = v^2
This shows that the velocity (v) at the top of the loop is the same as the velocity just before the loop.
Now, let's substitute this velocity into the equation for potential energy:
gh = 0.7v^2
gh = 0.7(v^2)
As we know that v^2 = rg (from the centripetal force equation), substitute v^2 with rg:
gh = 0.7(rg)
h = 0.7r
Therefore, the height (h) from which the ball should be launched to make it through the loop without falling off the track is 0.7 times the radius (r) of the ball.