Question

A car accelerates uniformly from rest and reaches a speed of 21.0 m/s in 9.05 s. Assume the diameter of a tire is 59.0 cm.

(a) Find the number of revolutions the tire makes during this
motion, assuming that no slipping occurs.

rev

(b) What is the final angular speed of a tire in revolutions per
second?

rev/s

Answer #1

v = u +a*t

a= (v-u) / t = ( 21-0) /9.05 =**2.32044199
m/s^2**

Calculate the Displacement.

Write the expression for Newton’s second equation of motion.

S = ut+0.5*a*t^2

= 0 +0.5***2.32*9.05^2 =**95.0069 m

Calculate the circumference of wheel.

Write the expression for circumference of wheel.

C = pi****d*

Here, *C* is the circumference and *d* is the
diameter of the wheel.

C = pi *59*10^-2 =1.85353967

Calculate the number of revolutions.

number of revlution

N = S/C = 95.0069/1.8535 =51.2581063

**The number of revolutions is 51.2581063**

b) v = r*omega

omega= V/r = 21 / ( 0.59/2) =71.1864407 rad/s

**=11.32967 rev/s**

let me know in comment if there is any problem or doubt

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