A torque of 60 N*m acts on a wheel of moment of inertia 30 kg*m^2 for 5 s and then is removed.
a) What is the angular acceleration of the wheel?
rad/s^2
b) How many revolutions does it make in 15 s if it starts at rest?
torque=net
To find the answers to these questions, we can use the formulas related to torque and angular motion. Let's start with part (a).
a) To find the angular acceleration of the wheel, we can use the formula:
Torque (τ) = Moment of inertia (I) * Angular acceleration (α)
Rearranging the formula, we can solve for α:
α = τ / I
Given that the torque (τ) is 60 N*m and the moment of inertia (I) is 30 kg*m^2, we can substitute the values into the formula:
α = 60 N*m / 30 kg*m^2
Simplifying the expression, we find:
α = 2 rad/s^2
Therefore, the angular acceleration of the wheel is 2 rad/s^2.
b) To determine the number of revolutions the wheel makes in 15 seconds, we need to consider the relationship between angular displacement, angular velocity, and time.
The formula to calculate angular displacement is:
θ = ωinitial * t + (1/2) * α * t^2
Given that the wheel starts at rest, the initial angular velocity (ωinitial) is 0.
The formula simplifies to:
θ = (1/2) * α * t^2
Now, let's calculate the angular displacement:
θ = (1/2) * 2 rad/s^2 * (15 s)^2
Simplifying further:
θ = 30 rad
Since 1 revolution is equal to 2π radians, we can convert the angular displacement to revolutions:
Number of revolutions = θ / (2π)
Number of revolutions = 30 rad / (2π)
Using a calculator, we find that the number of revolutions in 15 seconds is approximately 4.77 revolutions.
Therefore, the wheel makes approximately 4.77 revolutions in 15 seconds if it starts at rest.