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?
To find the magnitude of the yo-yo's linear acceleration, we can use the concept of rotational motion and relate it to linear motion.
The rotational inertia of the yo-yo is given as 760 g·cm^2. However, to use this value in our calculations, we need to convert it to standard SI units (kilograms and meters).
Given:
Rotational inertia (I) = 760 g·cm^2
Mass (m) = 200 g = 0.2 kg
To convert the rotational inertia to SI units:
1 g = 0.001 kg
1 cm^2 = (0.01 m)^2 = 0.0001 m^2
Therefore, the rotational inertia in SI units is:
I = 760 g·cm^2 = 760 * 0.001 kg * 0.0001 m^2 = 0.076 kg·m^2
Next, to determine the linear acceleration, we can relate it to the angular acceleration using the following equation:
α (angular acceleration) = a (linear acceleration) / r (radius)
The radius (r) is given as 4.0 mm, which we can convert to meters:
4.0 mm = 4.0 * 0.001 m = 0.004 m
Now we can rearrange the equation to solve for the linear acceleration:
a = α * r
To find α, we can use the concept of torque. The torque (τ) acting on the yo-yo is given by the equation:
τ = I * α
Rearranging the equation, we can solve for α:
α = τ / I
The torque acting on the yo-yo is due to its weight, which can be calculated as the product of its mass (m) and the acceleration due to gravity (g):
τ = m * g * r
Substituting the given values:
τ = 0.2 kg * 9.8 m/s^2 * 0.004 m = 0.00784 N·m
Now we can substitute the values of τ and I into the equation for α:
α = τ / I = 0.00784 N·m / 0.076 kg·m^2 = 0.1032 rad/s^2
Finally, substitute the value of α and r into the equation for a:
a = α * r = 0.1032 rad/s^2 * 0.004 m = 0.0004128 m/s^2
So, the magnitude of the yo-yo's linear acceleration is 0.0004128 m/s^2.