A yo-yo is made of two solid cylindrical disks, each of mass 0.053 kg and diameter 0.073 m, joined by a (concentric) thin solid cylindrical hub of mass 0.0052 kg and diameter 0.012 m. Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest.
The decrease in potential energy equals the increase in kinetic energy, most of which is rotational KE. First compute the moment of inertia of the yo-yo, I.
If it falls a distance L (which is 1.1 m in this case) then
M g L = (1/2) I w^2 + (1/2) M V^2
The angular rotation rate w can be replaced by V/r, where r is the radius of the cylindrical hub. Then solve for V.
V^2[1 + I/(Mr^2)] = 2*g*L
Take it from there.
I still don't get it.Moment of inertia of the yo-yo is 1/2 mr^2: 1/2(.053kg)(.0365^2) + 1/2(.0052kg)(.006^2)
which equals to 3.54*10^-5
so i plugged that into = square root of (2*9.8*1.1)/ (1 + (3.54*10^-5/(mr^2))
and i got something like 4.6 which is not a correct answer. Can you help me clarify this question please?
To calculate the linear speed of the yo-yo when it reaches the end of its string using conservation of energy, we need to consider the potential energy and kinetic energy of the system.
1. Calculate the potential energy:
The yo-yo is released from rest, so it starts with no kinetic energy. Therefore, all the potential energy is converted into kinetic energy at the end of the string.
The potential energy formula is: PE = mgh, where m is the total mass and h is the height.
In this case, the height h is equal to the length of the string, which is 1.1m.
Total mass (m) = mass of disks + mass of hub
= (2 * 0.053 kg) + 0.0052 kg
= 0.1112 kg
Potential energy (PE) = m * g * h
= 0.1112 kg * 9.8 m/s^2 * 1.1 m
= 1.1002 J
2. Calculate the kinetic energy:
The kinetic energy formula is: KE = 0.5 * m * v^2, where m is the total mass and v is the linear speed.
Since all the potential energy is converted into kinetic energy at the end of the string, we can equate the two:
PE = KE
1.1002 J = 0.5 * (0.1112 kg) * v^2
Solving for v:
v^2 = (2 * 1.1002 J) / 0.1112 kg
v^2 = 19.798 J/kg
v = √(19.798 J/kg)
v ≈ 4.45 m/s
Therefore, the linear speed of the yo-yo when it reaches the end of its 1.1 m long string is approximately 4.45 m/s.
To calculate the linear speed of the yo-yo when it reaches the end of its string, we can use the principle of conservation of energy. The yo-yo starts at rest, so its initial kinetic energy is zero. As it falls, its gravitational potential energy is converted into kinetic energy.
The gravitational potential energy (U) of the yo-yo is given by the equation U = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the height is equal to the length of the string (1.1 m). The mass (m) of the yo-yo is the sum of the masses of the disks and the hub: m = 2(0.053 kg) + 0.0052 kg = 0.1112 kg.
Therefore, the initial potential energy (U) of the yo-yo is U = (0.1112 kg)(9.8 m/s^2)(1.1 m) = 1.10696 J.
The final kinetic energy (K) of the yo-yo when it reaches the end of the string is equal to the initial potential energy. The equation for kinetic energy is K = 1/2 mv^2, where m is the mass and v is the linear velocity.
Therefore, 1.10696 J = (1/2)(0.1112 kg)(v^2).
To find the linear speed (v), we can rearrange the equation:
v^2 = (2)(1.10696 J) / (0.1112 kg)
v^2 = 19.926 J/kg
Taking the square root of both sides gives:
v = √(19.926 J/kg)
v ≈ 4.464 m/s
Therefore, the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest, is approximately 4.464 m/s.