A mass m = 76.0 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.7 m and finally a flat straight section at the same height as the center of the loop (19.7 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)
What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track?
To determine the minimum speed the block must have at the top of the loop, we can use the principle of conservation of energy.
1. Find the potential energy at the top of the loop:
The potential energy (PE) at the top of the loop is equal to the gravitational potential energy (mgh), where m is the mass (76.0 kg) and h is the height (19.7 m). Therefore, PE = mgh.
2. Find the minimum kinetic energy required at the top of the loop:
At the top of the loop, the block must have enough kinetic energy to counteract gravity and still maintain circular motion. The minimum kinetic energy required is equal to the sum of potential energy at the top of the loop and the gravitational potential energy lost on the way down to the loop.
Let v be the velocity of the block at the top of the loop. The potential energy lost on the way down to the loop can be calculated using the equation PE = mgh. The height difference is twice the radius of the loop (2R).
So, the minimum kinetic energy at the top of the loop is equal to PE + mgh.
Therefore, Kinetic Energy (KE) = (2mgh) + mgh = 3mgh.
3. Set the minimum kinetic energy equal to the kinetic energy at the top of the loop:
The kinetic energy at the top of the loop is given by the formula KE = (1/2)mv², where v is the velocity at the top of the loop.
Setting the two equations equal to each other, we have:
3mgh = (1/2)mv²
4. Solve for the minimum speed (v):
Rearranging the equation, we get:
v² = 6gh
v = √(6gh)
Substituting the known values, with g as the acceleration due to gravity (9.8 m/s²), we have:
v = √(6 * 9.8 * 19.7)
Calculating the value:
v ≈ 24.8 m/s
Therefore, the minimum speed the block must have at the top of the loop to make it around without leaving the track is approximately 24.8 m/s.