Question

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?

Answer 1

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

b) angular acceleratio is 0.854rad/sec^2

Step-by-step 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

`J1 = (1/12) * 6.9 * 4.88^2 = 13.693 kg m^2`

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

`J1 = (2/5) * 34.5 * 1.22^2 = 20.539 kg m^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
`J_1'=J_1 + m_1* d_1^2`

`J_1' = 13.693 + 2.44^2* 6.9 = 54.77 kg m^2`

for the sphere we get
`J_2' = J_2 + m_2* d_2^2`

`J2' = 20.539 + 34.5* 6.1^2 = 1304.28 kg^m2`

And the total moment of inertia for the first case is

`J_(t1) = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2`

b) F=476 N

The torque for system is given as

`M = F* d* sin(a)`

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree

`M = F* L/2`

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

`a_1 = (M)/(J_(t1))`

`= (1161.44)/(1359.05)`

= 0.854 rad/sec^2