Answer:
Explanation:
Given that,
Slender rod
Length of rod=80cm=0.8m
Mass of slender rod=0.12kg
Sphere Bob at one end
Mass M1=0.02kg
Sphere Bod at the other end
Mass M2 =0.05kg
Linear speed of mass 2 at the lowest point
We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.
So, at the lowest point,
∆U = M2•g•y2 + M1•g•y1
Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is -40cm(-04m) from the midpoint
∆U = M2•g•y2 + M1•g•y1
∆U=0.05•9.81•(-0.4) + 0.02•9.82•0.4
∆U=-0.1962+0.07848
∆U=-0.11772 Nm
Now, the moment of inertia of the rod is given as
I=∫r²dm
dm=2pdr
I= 2p∫r²dr
I= 2 × 0.12/0.8 ∫r²dr; from r=0 to 0.4
I=0.3 [r³/3] from r=0 to 0.4
I= 0.3 [ 0.4³/3 -0] ,from r=0 to 0.4
I=0.3 × 0.02133
I=0.0064kg/m².
calculating of inertia of the end masses.
I(1+2)=Σmr² = (m1+m2)r²
I(1+2)=(0.02+0.05)0.4²
I(1+2)=0.07×0.4²
I(1+2)=0.0112 kg/m²
Now, the Energy of the masses due to angular velocity is given as
K.E=½ (I + I(1+2))w²
K.E=½(0.0064+0.0112)w²
K.E= 0.0088w²
Using conservation of energy
The potential energy is equal to the kinetic energy of the system
K.E=P.E
0.0088w²=0.11772
Then, w²=0.11772/0.0088
w²=13.377
w=√13.377
w=3.66rad/s
Then, the relationship between linear velocity and angular velocity is given by
v=wr
v=3.66×0.4
v=1.463m/s
The required linear speed is 1.46m/s approximately
