A motorcycle accelerates uniformly from rest and reaches a linear speed of 33.4 m/s in a time of 11.6 s. The radius of each tire is 0.265 m. What is the magnitude of the angular acceleration of each tire?
Circumference=pi*2r = 3.14 * 0.53=1.665
m.
Va = 33.4m/s * 6.28rad/1.665m=126 rad/s
a=(V-Vo)/t = (126-0)/11.6=10.87 rad/s^2.
To find the magnitude of the angular acceleration of each tire, we need to use the formula relating linear acceleration and angular acceleration for objects in circular motion.
The linear acceleration can be determined using the formula for uniformly accelerated motion:
a = (v - u) / t
where:
a = linear acceleration
v = final linear speed
u = initial linear speed (which is 0 m/s since the motorcycle starts from rest)
t = time taken to reach the final speed
Substituting the given values:
v = 33.4 m/s
u = 0 m/s
t = 11.6 s
a = (33.4 - 0) / 11.6
a = 2.879 m/s^2
Next, we can relate linear acceleration to angular acceleration using the formula:
a = r * α
where:
a = linear acceleration
r = radius of the tire
α = angular acceleration
Substituting the known values:
a = 2.879 m/s^2
r = 0.265 m
2.879 m/s^2 = 0.265 m * α
Rearranging the equation, we can solve for α:
α = a / r
α = 2.879 m/s^2 / 0.265 m
α ≈ 10.885 rad/s^2
Therefore, the magnitude of the angular acceleration of each tire is approximately 10.885 rad/s^2.