a yo-yo of mass 440kg measuring 0.9m in radius was dropped from a platform 63m high. One end of the string was tied to the platform, so the yo-yo unwinds as it descends. Assuming the axle of the yo-yo has a radius of 0.2m, find the velocity of descent at the end of the fall. The acceleration of gravity is 9.81 m/s^2.
The velocity of the yo-yo at any height is r*w, where w is the angular velocity and r = 0.2 m.
Assume the yo-yo is a uniform disc, or pair of discs. Its moment of inertia is (1/2) M R^2, except for a small contribution for the axle, which we will have to ignore since they don't tell you its mass or length.
After dropping a distance H, the potential energy loss equals the kinetic energy gain.
M g H = (1/2)M V^2 + (1/2) I w^2
= (1/2)M V^2 + (1/4)M R^2*(V/r)^2
The mass M cancels out.
V^2 = g*H /[1/2 + (1/4)(R/r)^2]
THANK YOU SO MUCH!!! :)
To find the velocity of descent at the end of the fall, we can use the principle of conservation of mechanical energy.
The total mechanical energy of the yo-yo at the top platform is the sum of its potential energy and kinetic energy.
Potential energy at the top platform:
PE_top = m * g * h
= 440 kg * 9.81 m/s^2 * 63 m
= 271,644.6 J
Kinetic energy at the end of the fall:
KE_end = (1/2) * I * ω^2
= (1/2) * (m * r^2) * ω^2
= (1/2) * (440 kg * 0.2 m^2) * ω^2
= 8.8 kg.m^2 * ω^2
Since the yo-yo unwinds, the linear speed of descent (v) at the end of the fall is related to its angular velocity (ω) by the equation:
v = ω * r
= ω * (0.2 m)
Also, the angular velocity ω can be calculated using the concept of conservation of mechanical energy.
Total mechanical energy at the top platform:
E_top = PE_top + KE_top
The total mechanical energy at the end of the fall is the sum of its potential and kinetic energy:
E_end = PE_end + KE_end
= 0 + KE_end (no potential energy at the end of the fall)
Since mechanical energy is conserved:
E_top = E_end
Therefore:
PE_top + KE_top = KE_end
Substituting the values:
271,644.6 J + 0 = 8.8 kg.m^2 * ω^2
Simplifying:
8.8 kg.m^2 * ω^2 = 271,644.6 J
Dividing both sides by 8.8 kg.m^2:
ω^2 = 271,644.6 J / 8.8 kg.m^2
Calculating ω:
ω = sqrt(271,644.6 J / 8.8 kg.m^2)
= sqrt(30894.56)
≈ 175.89 rad/s
To find the linear speed of descent:
v = ω * r
= 175.89 rad/s * 0.2 m
≈ 35.18 m/s
Therefore, the velocity of descent at the end of the fall is approximately 35.18 m/s.
To find the velocity of descent at the end of the fall, we can use the principle of conservation of energy.
First, let's calculate the gravitational potential energy (GPE) of the yo-yo at the starting height:
GPE = mgh
Where:
m = mass of the yo-yo = 440 kg
g = acceleration due to gravity = 9.81 m/s^2
h = height = 63 m
GPE = 440 kg * 9.81 m/s^2 * 63 m
GPE = 273,356.4 Joules
Next, let's calculate the kinetic energy (KE) of the yo-yo at the end of the fall:
KE = 1/2 * mv^2
Where:
m = mass of the yo-yo = 440 kg
v = velocity at the end of the fall (what we want to find)
Since there is no potential energy left at the end of the fall (all converted to kinetic energy), we can equate the GPE and KE:
GPE = KE
mgh = 1/2 * mv^2
440 kg * 9.81 m/s^2 * 63 m = 1/2 * 440 kg * v^2
(440 kg * 9.81 m/s^2 * 63 m) / (1/2 * 440 kg) = v^2
v^2 = (440 kg * 9.81 m/s^2 * 63 m) / (1/2 * 440 kg)
v^2 = 9.81 m/s^2 * 63 m * 2
v^2 = 1236.54 m^2/s^2
Finally, we can take the square root of both sides to find the velocity:
v = √(1236.54 m^2/s^2)
v ≈ 35.15 m/s
Therefore, the velocity of descent at the end of the fall is approximately 35.15 m/s.