A yo-yo has a rotational inertia of 810 g·cm2 and a mass of 95 g. Its axle radius is 2.6 mm and its string is 100 cm long. The yo-yo is thrown so that its its initial speed down the string is 1.0 m/s.
(a) How long does the yo-yo take to reach the end of the string?
s
(b) As it reaches the end of the string, what is its total kinetic energy?
J
(c) As it reaches the end of the string, what is its translational speed?
m/s
(d) As it reaches the end of the string, what is its translational kinetic energy?
J
(e) As it reaches the end of the string, what is its rotational speed?
rad/s
(f) As it reaches the end of the string, what is its rotational kinetic energy?
J
Hint: Use conservation of energy, and the fact that the yo-yo is thrown DOWN. Rotational KE is (1/2) I w^2 and
V = r w
where w is the angular velocity and r is the axle radius.
To solve this problem, we'll need to use several concepts related to rotational dynamics and energy.
(a) To find the time it takes for the yo-yo to reach the end of the string, we can use the equation:
time = distance / speed
The distance traveled by the yo-yo is the length of the string, which is given as 100 cm. The initial speed down the string is given as 1.0 m/s. To make sure the units are consistent, we need to convert the length to meters. Therefore, the distance is 1.00 m.
Plugging in the values into the equation, we get:
time = 1.00 m / 1.0 m/s = 1.00 s
So, the yo-yo takes 1.00 second to reach the end of the string.
(b) To find the total kinetic energy of the yo-yo as it reaches the end of the string, we need to consider both its translational and rotational kinetic energies. The formula to calculate kinetic energy is:
kinetic energy = 0.5 * mass * speed^2
The mass of the yo-yo is given as 95 g, which needs to be converted to kilograms. Therefore, the mass is 0.095 kg.
The translational speed is given as 1.0 m/s. Plugging in the values, we find:
translational kinetic energy = 0.5 * 0.095 kg * (1.0 m/s)^2 = 0.0475 J
(c) The translational speed of the yo-yo as it reaches the end of the string remains the same as its initial speed, which is 1.0 m/s.
(d) The translational kinetic energy of the yo-yo as it reaches the end of the string is the same as the answer in part (b), which is 0.0475 J.
(e) To find the rotational speed of the yo-yo as it reaches the end of the string, we can use the concept of conservation of angular momentum. The equation for angular momentum is:
angular momentum = rotational inertia * rotational speed
The rotational inertia of the yo-yo is given as 810 g·cm^2, which needs to be converted to kg·m^2. Therefore, the rotational inertia is 0.081 kg·m^2.
We're also given the radius of the yo-yo's axle as 2.6 mm, which needs to be converted to meters. Therefore, the radius is 0.0026 m.
Since the rotational inertia is equal to mass * radius^2, we can rearrange the formula to solve for mass:
mass = rotational inertia / radius^2
mass = 0.081 kg·m^2 / (0.0026 m)^2 = 11.237 kg
To find the rotational speed, we divide the angular momentum by the rotational inertia:
rotational speed = angular momentum / rotational inertia
Since we don't have information about the angular momentum, we cannot solve for the rotational speed.
(f) Similarly, since we don't have the rotational speed, we cannot calculate the rotational kinetic energy either.