An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length l = 4.88 m to a uniform sphere with mass ms = 34.5 kg and radius r = 1.22 m. note ms = 5mr and l = 4r.

1)

what is the moment of inertia of the object about an axis at the left end of the rod?

kg-m2

2)

if the object is fixed at the left end of the rod, what is the angular acceleration if a force f = 476 n is exerted perpendicular to the rod at the center of the rod?

a)  total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is  d1=0.5*L

By Steiner theorem

for the rod we get

for the sphere we get

And the total moment of inertia for the first case is

b) F=476 N

The torque for system is given as

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree  

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

      = 0.854 rad/sec^2

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.