A small block of mass 42 kg slides along a frictionless loop-the-loop track. The radius of the loop is 5 meters. At what height above the bottom of the track should the block be released from so that it just makes it through the loop-the-loop without losing contact with the track?
Use SI units in your answer.
h_{block} =
To find the height above the bottom of the track at which the block should be released to just make it through the loop-the-loop without losing contact with the track, we can use the principle of conservation of energy.
The energy at the highest point of the loop (when the block is just about to lose contact with the track) is purely gravitational potential energy (PE). At this point, the block is at rest, so it has no kinetic energy (KE).
The total mechanical energy (E) of the block at any point along the track is the sum of the gravitational potential energy and the kinetic energy. Mathematically, we can express this as:
E = PE + KE
The gravitational potential energy is given by the formula:
PE = m * g * h
where m is the mass of the block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the bottom of the track.
The kinetic energy is given by the formula:
KE = (1/2) * m * v^2
where v is the velocity of the block.
At the highest point of the loop, the velocity is minimum and is equal to zero. Therefore, the kinetic energy is also zero.
Since the total mechanical energy is conserved, we can equate the initial and final energies:
E_initial = E_final
At the initial point (when the block is released from a height h above the bottom of the track), the block has only gravitational potential energy, so:
E_initial = m * g * h
At the final point (the highest point of the loop), the block has only kinetic energy, which is zero in this case:
E_final = 0
Equating the initial and final energies, we have:
m * g * h = 0
Simplifying the equation, we get:
h = 0
From this equation, we can see that the height above the bottom of the track at which the block should be released is zero. This means that the block should be released at the bottom of the loop in order to just make it through without losing contact with the track.
To find the height above the bottom of the track at which the block should be released in order to just make it through the loop-the-loop without losing contact with the track, we need to consider the forces acting on the block at different positions along the track.
At the top of the loop, the block experiences a net downward force due to its weight and a net inward force due to the centripetal force required to keep it moving in a circular path.
At the bottom of the loop, the block experiences a net upward force due to its weight and a net inward force due to the centripetal force required to keep it moving in a circular path.
For the block to maintain contact with the track throughout the loop-the-loop, the net inward force at the top of the loop must be greater than or equal to the net upward force at the bottom of the loop.
The net inward force at the top of the loop is the difference between the downward force due to the block's weight and the centripetal force:
F_inward_top = m*g - F_centripetal = m*g - m*v^2/r
The net upward force at the bottom of the loop is the sum of the upward force due to the block's weight and the centripetal force:
F_upward_bottom = m*g + F_centripetal = m*g + m*v^2/r
Where:
m = mass of the block (42 kg)
g = acceleration due to gravity (9.8 m/s^2)
v = velocity of the block at the top of the loop (which we need to find)
r = radius of the loop (5 m)
At the top of the loop, the block just barely makes it through the loop when contact is lost, which means the net inward force is equal to zero:
F_inward_top = 0
Substituting the expressions for F_inward_top and F_upward_bottom:
0 = m*g - m*v^2/r
m*v^2/r = m*g
Simplifying:
v^2 = g*r
v = sqrt(g*r)
So, the velocity of the block at the top of the loop is given by:
v = sqrt(9.8 m/s^2 * 5 m) = 9.9 m/s
Now, we can find the height above the bottom of the track at which the block should be released in order to have this velocity at the top of the loop.
The total mechanical energy of the block is conserved throughout the motion. At the bottom of the loop, the block only has gravitational potential energy, and at the release point, the block only has gravitational potential energy and kinetic energy.
At the bottom of the loop:
Ep_bottom = m*g*h_bottom
At the release point:
Ep_release + Ek_release = m*g*h_release + 0.5*m*v^2
Since mechanical energy is conserved:
Ep_bottom = Ep_release + Ek_release
m*g*h_bottom = m*g*h_release + 0.5*m*v^2
Simplifying:
h_release = h_bottom + 0.5*v^2/g
h_release = 0 + 0.5*(9.9 m/s)^2 / 9.8 m/s^2
h_release = 0 + 0.5*9.9 m = 4.95 m
Therefore, the block should be released from a height of 4.95 meters above the bottom of the track in order to just make it through the loop-the-loop without losing contact with the track.