Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo-yo is 0.056 kg, its moment of inertia is 2.9 X 10^-5 kg X m^2 and the radius r of the axle the string wraps around is 0.0064 m. Through what height must the yo-yo fall for its linear speed to be 0.60 m/s?
mgh=1/2 mv^2+1/2 *I w^2
but w=v/r
mgh=1/2 mv^2+1/2 *I*v^2/r^2
YOu know m,g, I ,r solve for v
h= 0.25
To determine the height through which the yo-yo must fall, we can use the principle of conservation of mechanical energy. The initial potential energy of the yo-yo will be equal to the final kinetic energy of the yo-yo.
1. Determine the initial potential energy (PE_initial) of the yo-yo.
PE_initial = m * g * h
Where:
m = mass of the yo-yo = 0.056 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height through which the yo-yo falls (unknown)
2. Determine the final kinetic energy (KE_final) of the yo-yo.
KE_final = (1/2) * I * w^2
Where:
I = moment of inertia of the yo-yo = 2.9 X 10^-5 kg X m^2
w = angular velocity = v / r
Since the yo-yo is moving in a linear motion, we need to convert the angular velocity (w) to linear velocity (v).
v = w * r
Given that the linear speed (v) is 0.60 m/s, we can calculate the angular velocity (w).
w = v / r = 0.60 m/s / 0.0064 m
w = 93.75 rad/s (approximately)
3. Calculate the final kinetic energy (KE_final) using the angular velocity.
KE_final = (1/2) * I * w^2 = (1/2) * (2.9 X 10^-5 kg X m^2) * (93.75 rad/s)^2
4. Set the initial potential energy (PE_initial) equal to the final kinetic energy (KE_final) and solve for the height (h).
m * g * h = (1/2) * I * w^2
Substituting the given values:
(0.056 kg) * (9.8 m/s^2) * h = (1/2) * (2.9 X 10^-5 kg X m^2) * (93.75 rad/s)^2
Solving for h:
h = [(1/2) * (2.9 X 10^-5 kg X m^2) * (93.75 rad/s)^2] / [(0.056 kg) * (9.8 m/s^2)]
Calculating this equation will give us the height (h) through which the yo-yo must fall for its linear speed to be 0.60 m/s.
To find the height through which the yo-yo must fall for its linear speed to be 0.60 m/s, we can use the principle of conservation of mechanical energy. We will equate the potential energy lost by the yo-yo while falling to the kinetic energy gained.
The potential energy lost by the yo-yo is given by the formula:
Potential energy = mgh
where m is the mass of the yo-yo, g is the acceleration due to gravity, and h is the height.
The kinetic energy gained by the yo-yo is given by the formula:
Kinetic energy = (1/2)mv^2 + (1/2)Iω^2
where m is the mass of the yo-yo, v is the linear speed of the yo-yo, I is the moment of inertia of the yo-yo, and ω is the angular velocity of the yo-yo.
In this case, the yo-yo is dropped from rest, so its initial angular velocity is zero (ω = 0).
By equating the potential energy lost to the kinetic energy gained, we have:
mgh = (1/2)mv^2 + (1/2)Iω^2
Since ω = v/r (where r is the radius of the axle), we can substitute this expression into the equation:
mgh = (1/2)mv^2 + (1/2)I(v/r)^2
Now we can solve for h:
h = (1/2)v^2 + (1/2)(I/m)(v/r)^2
Substituting the given values:
m = 0.056 kg
I = 2.9 X 10^-5 kg X m^2
r = 0.0064 m
v = 0.60 m/s
h = (1/2)(0.60 m/s)^2 + (1/2)(2.9 X 10^-5 kg X m^2 / 0.056 kg)((0.60 m/s) / 0.0064 m)^2
Calculating this expression will give us the height h through which the yo-yo must fall for its linear speed to be 0.60 m/s.