a block slides down a frictionless ramp. at the bottom of the ramp is a vertical circular loop with a radius of 75 cm. from what minimum height does the block need to be released in order to make it around the loop?
let h be the height of the release.
Then h-2r is the altitude of the top of the loop.
Initial PE=finalPE+final KE
mgh=mg( 2r)+1/2 m v^2
but v^2/r= g to keep it at the top of the loop.
mgh= 2mgr+1/2 mg*r
h= 2r+1/2 r=2.5 r
check my math.
To determine the minimum height from which the block needs to be released to make it around the loop, we need to consider the forces acting on the block at different points.
First, let's analyze the forces acting on the block at the top of the loop. At this point, the block is momentarily at rest.
1. Gravitational force (mg): The weight of the block acts vertically downwards.
2. Normal force (N): The normal force acts perpendicular to the surface of the loop. At the topmost point of the loop, the normal force is equal to the centripetal force required to keep the block moving in a circular path.
Centripetal force (Fc): The centripetal force acts towards the center of the circular path and is given by the equation:
Fc = mv² / r
Where m is the mass of the block, v is its velocity, and r is the radius of the loop.
To determine the minimum height, we need to find the velocity of the block at the topmost point of the loop. Since the block is sliding down a frictionless ramp, we can use the conservation of mechanical energy to find its velocity.
Total mechanical energy (E) is conserved and is given by the sum of the potential energy (PE) and kinetic energy (KE) of the block:
E = PE + KE
At the top of the ramp, all of the block's potential energy is converted into kinetic energy.
PE = mgh
Where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the ramp.
KE = 1/2 mv²
Setting PE equal to KE, we can solve for the velocity (v):
mgh = 1/2 mv²
Canceling the mass (m) on both sides of the equation and solving for v:
v = √(2gh)
Now that we have the velocity of the block at the topmost point of the loop, we can equate the centripetal force to the normal force to find the minimum height.
Fc = N
mv² / r = N
Substituting the expression for v:
m(2gh) / r = N
Now we can solve for h:
h = (Nr) / (2g)
Using the given radius of the loop (75 cm = 0.75 m), we can plug in the necessary values to find the minimum height required for the block to make it around the loop.