physics and torque

Question

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During a very quick stop, a car decelerates at 6.8 m/s². Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement).

a_t = 6.8 m/s²

r = 0.255 m

ω₀ = 91 rad/s

a) what is the angular acceleration of its tires in rad/s^2, assuming they have a radius of 0.255 m and do not slip on the pavement?

b) how many revolutions do the tires make before coming to rest, given their initial angular velocity is 91 rad/s?

c) how long does the car take to stop completely in seconds?

d) what distance does the car travel in this time in meters?

e) what was the cars initial speed in m/s?

Answer 1

a) the equation for relating tangential acceleration and angular acceleration is a_t = αr. Therefore α = a_t/r α = (6.8m/s²)/0.255m = 26.67 rad/s²

b) I found the revolutions before rest by first finding the time the car takes to stop.

Using the equation: ω_f = ω₀ + αt, and knowing that the final angular velocity, ω_f is equal to zero when the car is stopped, we get 0 = (91rad/s) - (26.67rad/s²)t, and t = 3.41 seconds. From here we can use the equation: θ = ω₀t + 0.5αt² to find θ, or the number of radians the tires traveled before stopping. Using our calculated t value, θ = 91rad/s(3.41s) + 0.5(-26.67rad/s²)(3.41s)², θ = 155.2 radians. We know that 1 rotation is equal to 2pi radians, we simply divide θ by 2pi to get 24.7 rotations

c) As calculated in b), t = 3.41 seconds

d) The equation relating tangential distance and rotational distance is x_t = θr. Using our calculated θ value, x_t = 155.2rad(0.255m) = 39.6m

e) The equation relating tangential velocity and rotational velocity is v_t = ωr. Using the initial ω value given, we get v_t = 91rad/s(0.255m) = 23.2 m/s