A car travels at 80 km/h on a level road in the positive direction of an x axis.

Question

A car travels at 80 km/h on a level road in the positive direction of an x axis. Each tire has a diameter of 66 cm and mass 10kg. What are the velocity at the center, top, and bottom of the tire and The magnitude a of the acceleration at the center, top, and bottom of each tire? What are the velocity Ti at the center, top, and bottom of the tire and The magnitude a of the acceleration at the center, top, and bottom of each tire? What is the kinetic energy of each tire? Why is the kinetic energy different than for either simple linear motion or simple rotational motion? If a non-spinning wheel impacted an object, how would the subsequent motion of both objects be different from the case of a spinning wheel? How does this relate to the kinetic energy formula? How is the new model of kinetic energy a give and take type model? What kind of friction is important to rolling motion? Is the bottom of a tire stopped, either ideally or in reality? What can these questions tell us about the importance of anti-lock breaking systems?

Explanation / Answer

1. a) centre of tire = vp = 80 km/h Or (80x1000k / 3600sec) = 22.22 m/s

relative to woment sitting in car = 0

b) at top,

vp = v + wr = 2v = 2 x 80 = 160 km/h or 44.44 m/s

relative to 0 sitting in car = 160i - 80i = 80i km/h

c) vp = v - wr = 0

relatibe to women = 0 - 80i = - 80i km/h

2.a) there is no tangentical acc. so accc. = 0

a = 0

b) top point will feel centripetal acc.

at top a = v^2 / r = 22.22^2 / 0.33 = 1496.15 m/s^2 (-j)

downwards direction.

c) bottom of titr

a = 1496.15 j m/s^2

upwards direction

2.

3. a ) v = 80 i km/h

b) v = 160i km/h

c) v = 0 km/h

4. a) a = 0

b) - 1496.15j m/s^2

c) a = 1496.15 j m/s^2

1. KE = translational KE + rotational KE

      = mv^2 /2 + Iw^2 /2

if we consider tire as hoop then I = mr^2

so Ke = mv^2 /2 + (mr^2)(v/r)^2 /2 = mv^2

KE = 10 x 22.22^2 = 4937.28 J

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