Solve for b using a rotational kinematic equation:
a) Calculate the linear acceleration of a car, the 0.310-m radius tires of which have an angular acceleration of 11.5 rad/s2. Assume no slippage and give your answer in m/s2
b) How many revolutions do the tires make in 2.50 s if they start from rest?
a) We know that the linear acceleration of a point on a rotating object can be calculated using the equation:
a = r * α
where:
a = linear acceleration
r = radius of the tires = 0.310 m
α = angular acceleration = 11.5 rad/s^2
Plugging in the values:
a = 0.310 * 11.5
a = 3.565 m/s^2
Therefore, the linear acceleration of the car is 3.565 m/s^2.
b) To find the number of revolutions the tires make in 2.50 s, we first need to calculate the final angular velocity of the tires using the kinematic equation:
ω = ω₀ + α * t
where:
ω = final angular velocity
ω₀ = initial angular velocity (which is 0 since the tires start from rest)
α = angular acceleration = 11.5 rad/s^2
t = time = 2.50 s
Plugging in the values:
ω = 0 + 11.5 * 2.50
ω = 28.75 rad/s
Now, we can find the number of revolutions using the equation:
θ = ω₀ * t + 0.5 * α * t^2
where:
θ = total angle rotated in radians
ω₀ = initial angular velocity (0 rad/s)
α = angular acceleration = 11.5 rad/s^2
t = time = 2.50 s
Plugging in the values:
θ = 0 * 2.50 + 0.5 * 11.5 * (2.50)^2
θ = 0 + 0.5 * 11.5 * 6.25
θ = 0 + 35.9375
θ = 35.9375 radians
To convert this into revolutions, we know that 1 revolution is equal to 2π radians. Therefore:
Number of revolutions = θ / (2π)
Number of revolutions = 35.9375 / (2 * 3.14159)
Number of revolutions ≈ 5.72 revolutions
Therefore, the tires make approximately 5.72 revolutions in 2.50 s.