The small mass m sliding without friction along a looped track is to remain on the track at all times, even at the very top of the loop of radius r.
a) In terms of the given quantities, determine the minimum release height h
b) If the actual release height is 5h, calculate the normal force exerted by the track at the top of the loop
To determine the minimum release height (h) for the small mass to remain on the looped track, we need to consider the forces acting on the mass at the top of the loop.
a) At the top of the loop, the only two forces acting on the mass are gravity (mg) pointing downwards and the normal force (N) exerted by the track pointing towards the center of the loop.
To find the minimum release height, we need to set up the condition for the mass not to leave the track. This condition is that the net force on the mass at the top of the loop is equal to or greater than zero:
Net Force = Centripetal Force + Weight
The centripetal force is given by:
Centripetal Force = (mass) x (acceleration towards the center of the loop)
At the top of the loop, the acceleration towards the center is directed downwards, so we have:
Centripetal Force = (mass) x (acceleration due to gravity)
Therefore, the net force becomes:
Net Force = (mass) x (acceleration due to gravity) + Weight
We can express the weight as:
Weight = mass x gravity
Substituting this into the net force equation, we have:
Net Force = (mass) x (acceleration due to gravity) + (mass x gravity)
Since the net force should be zero or greater, we can set up the inequality:
0 ≤ (mass) x (acceleration due to gravity) + (mass x gravity)
We can rewrite this inequality in terms of the given quantities, mass (m) and radius of the loop (r):
0 ≤ m(g + a)
Where g is the acceleration due to gravity (g = 9.8 m/s²) and a is the acceleration towards the center of the loop (a = v²/r).
Since the mass cancels out on both sides of the inequality, we have:
0 ≤ g + a
Now, we can solve for the minimum release height (h) in terms of the radius (r):
a = v²/r
To determine the minimum velocity (v) required for the mass to remain on the track at the top of the loop, we can use conservation of energy. At the bottom of the loop, the gravitational potential energy is converted into kinetic energy, so:
mgh = 1/2 mv²
Where h is the release height, and v is the velocity at the top of the loop.
Rearranging this equation, we get:
v² = 2gh
Substituting this into the previous equation for a:
a = (2gh)/r
Now, we can substitute this value of a into the inequality:
0 ≤ g + (2gh)/r
Rearranging the inequality to solve for h, we have:
0 ≤ g(1 + 2h/r)
Dividing both sides by g gives:
0 ≤ 1 + 2h/r
Subtracting 1 from both sides:
-1 ≤ 2h/r
Finally, solving for h:
h ≥ -r/2
Therefore, the minimum release height (h) is equal to or greater than -r/2.
(Note: This means the release height should be at least half of the radius of the loop in order for the mass to remain on the track at all times.)
b) If the actual release height is 5h, we can substitute this value into the expression for the normal force at the top of the loop.
The normal force (N) at the top of the loop can be calculated as:
N = mg + mv²/r
Substituting h = 5h:
N = mg + m(2gh)/r
N = mg + 2mgh/r
Now, we can substitute the values of g and v² from earlier:
N = mg + 2m(2gh)/r
N = mg + 4mgh/r
Simplifying:
N = mg(1 + 4h/r)
Finally, we can substitute the given value of h = 5h:
N = mg(1 + 4(5h)/r)
N = mg(1 + 20h/r)
Therefore, the normal force exerted by the track at the top of the loop is mg(1 + 20h/r).