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

Answer:

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

$J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2$

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

$J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2$

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 $J_1'=J_1 + m_1\times d_1^2$

$J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2$

for the sphere we get $J_2' = J_2 + m_2\times d_2^2$

$J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2$

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

$M = F\times d\times sin(a)$

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree  

$M = F\times L/2$

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

$a_1 = \frac{M}{J_{t1}}$

      $= \frac{1161.44}{1359.05}$

      = 0.854 rad/sec^2