The figure below gives the speed v versus time t for a 0.510 kg object of radius 5.90 cm that rolls smoothly down a 30° ramp. What is the rotational inertia of the object?
1 kg·m2
To find the rotational inertia of the object, we need to use the relationship between the angular acceleration (α) and the torque (τ) acting on the object. The formula is τ = Iα, where τ is the torque and I is the rotational inertia.
In this case, since the object is rolling smoothly down the ramp, there is no slipping. This means that the linear acceleration (a) of the object is related to the angular acceleration (α) by the equation a = Rα, where R is the radius of the object.
Given that the radius of the object is 5.90 cm, we need to convert it to meters:
R = 5.90 cm = 0.0590 m
The ramp angle is given as 30°. We can use this information to find the linear acceleration (a) of the object using the equation a = gsinθ, where g is the acceleration due to gravity (approximately 9.8 m/s²) and θ is the ramp angle in radians. To convert degrees to radians, we use the formula θ (rad) = (π/180) × θ.
θ = 30° = (π/180) × 30 = π/6 radians
a = gsinθ = (9.8 m/s²) × sin(π/6) ≈ 4.9 m/s²
Now, we know that a = Rα, so we can rearrange the equation to solve for α:
α = a / R = (4.9 m/s²) / (0.0590 m) ≈ 83.05 rad/s²
Finally, we can use the given torque (τ) and the calculated α to find the rotational inertia (I). The formula is τ = Iα, so we can rearrange it to solve for I:
I = τ / α
Since the question doesn't provide a specific torque value, we cannot calculate the exact rotational inertia (I). However, given the provided options, 1 kg·m² is the closest value to the actual solution.
So, the rotational inertia of the object is approximately 1 kg·m².