A 500 g block slides down a frictionless inclined plane. At the bottom, a vertical circular loop with radius 67 cm is in the block’s path. From what height would the block need to be released to successfully complete the loop without losing contact with the surface of the track?
What i did is:
mv^2/r=T+mg
v=sr(T+mg)r/m
But i don't know what T is, help.
To solve for the required height, you need to consider the forces acting on the block at the top of the loop. At that point, the block is experiencing both gravity (mg) and the normal force (N) from the circular track.
The normal force, N, acts perpendicular to the surface of the track and provides the necessary centripetal force to keep the block in circular motion. The tension, T, in the track acts in the opposite direction of gravity to balance it out.
Since the block is not losing contact with the surface, the normal force, N, needs to be greater than or equal to zero. In other words, the minimum requirement for the block to remain on the track is N ≥ 0.
At the top of the loop, the normal force N is equal to the sum of the tension T and the gravitational force mg. Therefore, we have:
N = T + mg
Now, let's consider the relationship between velocity (v), radius (r), and acceleration (a) at the top of the loop. The centripetal acceleration is given by a = v^2/r.
Since a = g at the top of the loop (both the gravitational acceleration and the centripetal acceleration point downwards), we have:
g = v^2/r
Solving for v in terms of g and r:
v = √(g * r)
Now, substitute this expression for v back into the equation for N at the top of the loop:
N = T + mg = mv^2 / r
Substituting v = √(g * r):
N = T + mg = m(g * r) / r = mg
We conclude that N = mg. Therefore, the tension T in the track is zero at the top of the loop.
Now, let's analyze the block's condition at the top of the loop. At that point, the block momentarily loses contact with the track, and the only force acting on it is gravity (mg). For this to happen, the block's weight should be equal to or greater than the centripetal force required to maintain circular motion:
mg ≥ mv^2 / r
Rearranging the equation:
mg ≥ m(g * r) / r
Simplifying:
g ≥ g
Since g is always equal to g, this means that the block will momentarily lose contact with the track. In order for the block to successfully complete the loop without losing contact, the inequality equation needs to hold strictly, not just be equal.
Therefore, there is no height from which the block can be released to complete the loop without losing contact with the surface of the track.