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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards