an automobile tire has a radius of 0.350M, and its center forward with a liner speed of v=13.0m/s.
(a) Determine the angular speed of the wheel. (Assume that there is no slipping of the surfaces in contact during the rolling motion.)
(b) Relative to the axle, what is the tangential speed of a point located 0.175m from the axle?
Did you mean to write:
Its center MOVES forward at a LINEAR speed of 13.0 m/s"
(a)Let w be the angular roation speed of the wheel, abd R its radius. The linear speed of the center of a nonslipping wheel is given by
v = R w. Solve for w
(b) For any point on the wheel with radius r, the tangential speed is
v' = r w.
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(a) To determine the angular speed of the wheel, we use the formula v = Rω, where v is the linear speed of the center of the wheel, R is its radius, and ω is the angular speed.
Given that the radius (R) of the tire is 0.350 m and the linear speed (v) is 13.0 m/s, we can rearrange the formula to solve for ω:
ω = v / R
Substituting the values, we get:
ω = 13.0 m/s / 0.350 m
ω ≈ 37.1429 rad/s (rounded to 4 decimal places)
Therefore, the angular speed of the wheel is approximately 37.1429 rad/s.
(b) To find the tangential speed of a point located 0.175 m from the axle, we use the formula v' = rw, where v' is the tangential speed, r is the distance from the axle, and w is the angular speed.
Given that the distance from the axle (r) is 0.175 m, and we already determined the angular speed (w) to be 37.1429 rad/s, we can calculate v':
v' = 0.175 m × 37.1429 rad/s
v' ≈ 6.495 m/s (rounded to 3 decimal places)
Therefore, the tangential speed of a point located 0.175 m from the axle is approximately 6.495 m/s.