Q1. The crankshaft in a race car goes from rest to 3100rpm in 2.8s.What is the c

Question

Q1. The crankshaft in a race car goes from rest to 3100rpm in 2.8s.What is the crankshaft's angular acceleration?

Q2. A frictionless pulley, which can be modeled as a 0.83kg solid cylinder with a 0.30m radius, has a rope going over it, The tensions in the rope are 12N and 10N . What is the angular acceleration of the pulley?

Q3. A 1.2g pebble is stuck in a tread of a 0.73m -diameter automobile tire, held in place by static friction that can be at most 3.3N . The car starts from rest and gradually accelerates on a straight road. How fast is the car moving when the pebble flies out of the tire tread?

Explanation / Answer

(1)

Change rpm to radians/s; then find the angular acceleration

3100rpm =3100r/m
3100r/m x 1m/60s x (2pi)rad/r = [6200(pi)/60]rad/s = 103.3pi rad/s

angular acc. =(103.3pi rad/s)/2.8s

angular acc.=116 rad/s^2

(2)

the difference in tensions in the pulley causes a net torque

the net torque is (T2-T1) R = 2 N x 0.3 m = 0.6 Nm

the torque on an object causes an angular acceleration given by

torque = I alpha where I is the moment of inertia and alpha the angular acceleration

the moment of inertia of a cylinder is 1/2 MR^2, so for this pulley

I = 1/2 (0.83kg)(0.3m)^2

I= 0.03735

finally, ang accel= net torque/I

=0.6/0.03735

= 16.06 rad/s/s

(3)

The maximum static friction (3.3 N) is your centripetal force, is the force pulling the pebble, making it stay.

F = ma, and for the centripetal force F = mv^2/r.
Your maximum F is 3.3 N, mass and radius can't be changed,

The transition takes place at:
v = sqrt(F*r/m)

v= sqrt((3.3*0.73)/0.0012)

v= 44.80m/s

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