Respuesta :
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
[tex]J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2[/tex]
moment of inertia for sphere is given as
J1=(2/5) ×m×r^2
[tex]J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2[/tex]
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 [tex]J_1'=J_1 + m_1\times d_1^2[/tex]
[tex]J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2[/tex]
for the sphere we get [tex]J_2' = J_2 + m_2\times d_2^2[/tex]
[tex]J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2[/tex]
And the total moment of inertia for the first case is
[tex]J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2[/tex]
b) F=476 N
The torque for system is given as
[tex]M = F\times d\times sin(a)[/tex]
where a is angle between Force and distance d
and where d represent distance from rotating axis.
In this case a = 90 degree
[tex]M = F\times L/2[/tex]
M=476*2.44 = 1161.44 Nm
The acceleration is calculated as
[tex]a_1 = \frac{M}{J_{t1}}[/tex]
[tex]= \frac{1161.44}{1359.05}[/tex]
= 0.854 rad/sec^2
