2 physics questions please help?

Question

1) A thin rod length=1.5m is oriented vertically with its bottom end attached to the floor by means of a frictionless hinge. the mass of the rod may be ignored compared to the mass of an object fixed to the top of the rod. the rod starting from rest tips over and rotates downward. what is the angular speed of the rod just before it strikes the floor? what is the magnitude of the angular acceleration of the rod just before it hits the floor?

2)The front and rear sprockets on a bike have a radii of 9.00 and 5.10 cm. the angular speed of the front sprocket is 9.4 rad/s. what is the linear speed in cm/s of the chain as if moves between the sprockets and the centripetal acceleration in cm/s^2 of the chain as it passes around the rear sprocket?

please show your work and god bless!

Answer 1

For the first question, we can start by looking at conservation of energy of the mass falling through the distance of the length of the rod to find the tangential velocity.
Since the motion will be exactly vertical at the moment it touches the floor
.5*m*v^2=m*g*h
v=sqrt(2*g*h)

since h=R=1.5
v=sqrt(2*g*R)

w, or angular speed is equal to v/R

in this case
w=R*sqrt(2*g*R)
=1.5*SQRT(2*9.81*1.5)
=8.14 rad/s

the acceleration is related to the torque as
T=a*m
since T=m*g*R
a=g*R
=1.5*9.81
=14.7 rad/s^2

The second question involves the relationship of two pulleys essentially. Note that the v for the chain is constant.

So w=v/R
v=w*R
for the front
v=9.4*9.00
=84.6 cm/sec
(BTW the front sprocket is called a chainring)

The rear sprocket, also called a cog, has R=5.10

w=v/R
w=84.6/5.1

w=16.588

also note that
wcog=wchainring*(Rchainring/Rcog)

the centripetal acceleration is
a=w^2*R
=16.588*5.1
=84.6

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