θ=2+4t^3；

So ω = θ' =12t2;

β=θ" = 24t；

So at =βr = 2.4t；;

So the speed = 4.8m/s.

The size of at is half that of A, which means that centripetal acceleration A ρ = √ 3at = √ 32.4t.

And aρ=ω2r = 14.4t 4.

So t 3 = 5/√ 3,

So θ = 2+4t 3 = 2+20/√ 3.

2)

Because it is a completely inelastic collision: V'=mV/(M+m),

According to the law of conservation of energy: μ (m+m) GL = m 2v 2/2 (m+m)

So l = m 2v 2/[2 μ (m+m) 2g]

3)

β=(ω-ω0)/(t-t0)=-0.05rad/s^2

When the flywheel stops, the time is: t=|ω0/β|= 100s.

So θ = ω 0t+β t 2/2 = 250 rad.

4) The resultant torque is not necessarily 0: Take only two forces as an example: when the acting points of the two forces are different and the connecting lines of the acting points are not parallel to the forces, the following forces are formed, and the torque is obviously not 0.

↓

―――

........↑

Only momentum is conserved at this time. (Angular momentum is not conserved because angular velocity will change because the torque is not zero; Mechanical energy is not necessarily conserved, because the external force has done work. )

5)

The theorem of moment of momentum is that the total change of angular momentum of the system is equal to the accumulation of resultant torque with time.

dL=Mdt

The conservation of momentum is the conservation of angular momentum (angular momentum is constant), provided that the resultant torque is 0.

6)

e=ka^2/2,= 1.0j,a=0. 1m，

So k = 2e/a2 = 200 n/m.

And e = mv 2/2, v =1.0m/s.

So m=2kg,

So the frequency v = 1/2 π√ (k/m) = 5/π Hz.

7)

T=2π√(m/k)

k=2E/A^2，

So t = 2π √ (Ma 2/2e) = 2π √ (m/2e)

So: e = 4ma 2 π 2/2t.