A yo-yo has a rotational inertia of 760 g·cm2 and a mass of 200 g. Its axle radius is 4.0 mm, and its string is 100 cm long. The yo-yo rolls from rest down to the end of the string.
(a) What is the magnitude of its linear acceleration?
cm/s2
(b) How long does it take to reach the end of the string?
s
(c) As it reaches the end of the string, what is its linear speed?
cm/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 kinetic energy?
J
(f) As it reaches the end of the string, what is its angular speed?
rad/s
just part a...does anyone know how to do it?
To solve this problem, we'll need to use several formulas related to rotational motion and apply Newton's second law. Let's break it down step by step.
(a) To find the magnitude of the yo-yo's linear acceleration, we can use the formula:
α = a / r
where α is the angular acceleration, a is the linear acceleration, and r is the radius of the axle. We can rearrange this formula to solve for a:
a = α * r
The question provides the rotational inertia of the yo-yo as 760 g·cm^2. Rotational inertia is denoted by the symbol I and is given by:
I = m * r^2
where m is the mass and r is the radius of the axle. Since we know I and r, we can solve for m:
m = I / r^2
Substituting the given values, we have:
m = (760 g·cm^2) / (4.0 mm)^2
Converting millimeters to centimeters, we get:
m = (760 g·cm^2) / (0.4 cm)^2 = 19000 g
Converting grams to kilograms:
m = 19 kg
Now we can find the angular acceleration:
α = τ / I
where τ is the torque. In this case, the torque τ is caused by the force of gravity pulling the yo-yo downward. The torque τ is given by:
τ = m * g * r
where g is the acceleration due to gravity. Substituting the known values:
τ = (19 kg) * (9.8 m/s^2) * (0.4 cm) = 74.64 mN·m
Converting millinewton meters to newton meters:
τ = 0.07464 N·m
Now we can find the angular acceleration α:
α = (0.07464 N·m) / (760 g·cm^2)
Converting grams and centimeters to kilograms and meters:
α = (0.07464 N·m) / (0.0760 kg·m^2) = 0.98 rad/s^2
Finally, we can calculate the linear acceleration:
a = α * r = (0.98 rad/s^2) * (0.4 cm) = 0.392 cm/s^2
Therefore, the magnitude of the yo-yo's linear acceleration is 0.392 cm/s^2.
(b) To find the time it takes for the yo-yo to reach the end of the string, we can use the formula:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 since it starts from rest), a is the acceleration, and t is the time. Rearranging the formula to solve for t:
t = (v - u) / a = v / a
We know that the final velocity v is given by:
v = u + at
Since the initial velocity u is 0, the formula simplifies to:
v = at
Substituting the known values:
t = (100 cm/s) / (0.392 cm/s^2) = 255.10 s
Therefore, it takes approximately 255.10 seconds for the yo-yo to reach the end of the string.
(c) The linear speed of the yo-yo at the end of the string is given by:
v = u + at
Since the initial velocity u is 0, we have:
v = at
Substituting the known values:
v = (0.392 cm/s^2) * (255.10 s) = 100 cm/s
Therefore, the linear speed of the yo-yo at the end of the string is 100 cm/s.
(d) The translational kinetic energy (KE) is given by:
KE = (1/2) * m * v^2
Substituting the known values:
KE = (1/2) * (19 kg) * (100 cm/s)^2 = 95000 J
Therefore, the translational kinetic energy of the yo-yo at the end of the string is 95000 Joules.
(e) The rotational kinetic energy (KE_rot) is given by:
KE_rot = (1/2) * I * ω^2
where ω is the angular speed. To find ω, we can use the formula:
ω = α * t
Substituting the known values:
ω = (0.98 rad/s^2) * (255.10 s) = 250 rad/s
Now we can calculate the rotational kinetic energy:
KE_rot = (1/2) * (760 g·cm^2) * (250 rad/s)^2
Converting grams and centimeters to kilograms and meters:
KE_rot = (1/2) * (0.0760 kg·m^2) * (250 rad/s)^2 = 23750 J
Therefore, the rotational kinetic energy of the yo-yo at the end of the string is 23750 Joules.
(f) The angular speed ω is given by:
ω = α * t
Substituting the known values:
ω = (0.98 rad/s^2) * (255.10 s) = 250 rad/s
Therefore, the angular speed of the yo-yo at the end of the string is 250 rad/s.