Question

1) A solid ball of radius 7.25 cm and mass 7.43kg starts from rest and rolls without slipping down a 20.7 degree incline that is 1.22m long. Calculate the angular velocity (in rad/s) of the ball when it reaches the bottom

2:43 PM o 37 BD ooooo AT&T LTE 2)A uniform thin rod of length 0.452 m and mass 5.70 kg is suspended freely from one end. It is pulled to the side an angle 38.8 degrees and released. If friction can be ignored, what is the speed of its free end (in m/s), at the lowest point?

Answer 1

Q1.

mass=m=7.43 kg

radius=r=7.25 cm=0.0725 m

let final speed at the bottom be v m/s.

then final angular speed=v/r rad/s

moment of inertia of a solid sphere=0.4*mass*radius^2

balancing total mechanical energy:

initial potential energy=final linear kinetic energy + final rotational kinetic energy

==>mass*g*height=0.5*mass*v^2+0.5*moment of inertia*angular speed^2

==>mass*g*height=0.5*mass*v^2+0.5*0.4*mass*r^2*(v/r)^2

==>9.8*1.22*sin(20.7)=0.5*v^2+0.2*v^2=0.7*v^2

==>v=sqrt(9.8*1.22*sin(20.7)/0.7)=4.0517 m/s