Torque related question?Anyone please help

Two spheres each having a mass of 3kg are attached to a rod of length 0.8 m and of negligible mass.Determine the time t in which the torque M= (8t) N-m,where t is in seconds, must be applied to the rod so that each sphere attains a speed of 3m/s starting from rest.(Rod rotates about its centre)

we know torque=MIxangular acceleration
=MIx a/r wherw a=linear acceleration, r=radius.

so here 8t = MI/r . dv/dt
or 8tdt=MI/r dv
or 8t^2/2= MI/r v
or 4t^2=2mr^2/r v
or 2t^2= mrv as v=0 at t=0;


or t=root(mrv/2)
=root(3x0.4x3/2)
=root1.8
=1.34sec

so required time is 1.34 sec.

The rod/sphere system will accelerate with angular acceleration α, where

α = T/I

T = applied torque, and I the moment of inertia of the system about the rod center

The angular velocity ωf at time tf is

ωf = ∫α*dt [t = 0 to t = tf]

ωf = ∫T/I*dt = (1/I)*∫8*t*dt = (1/I)*4*t² [0, tf] = 4*tf²/I

the angular velocity for the spheres moving at velocity v is v/R:
ωf = 3/0.4 = 7.5 s^-1

If the spheres' radius is small compared to 0.4 m then the moment of inertia is

I = 2*m*R² = 2*3*0.4² = 0.96 kg-m²

7.5 = 4*tf²/0.96 tf = √[7.5*0.96/4] = 1.3 s

M is torque, which is also the force that would be applied to a mass of 1 kg at 1 metre from the point where the torque is applied.

However, mass is 2 * 3 = 6 kg
and the radius is 0.8 / 2 = 0.4 m.

So a force of M / 0.4 has to accelerate a mass of 6 kg to 3 m/s

Force = mass * acceleration

8t / 0.4 = 6 * a

but v = u + at where v is the final velocity (3 m/s) and u is the initial velocity (0).
so
3 = at or a = 3/t

Therefore

8t / 0.4 = 6 * 3 / t

8t^2 = 7.2
t = root (7.2 / 8) = 0.95 seconds to 2 decimal places.

Moment of inertia of the system
= 2 * mr^2
= 2 * 3 * 0.4^2
= 0.96
torque = MoI * a
8t = 0.96 * a
where a = angular acceleration
a = 8.33333 t
sphere attains a speed of 3m/s
w =angular speed
3 = rw = 0.4 w
w = 7.5 rad/s
v =u+at
7.5 = 0 + 8.33333 t * t
t = 0.948s
answer