Answer:
Part a)
Mass of m2 is given as
Part b)
Angular acceleration is given as
Part c)
Tension in the rope is given as
Explanation:
Part a)
When m1 and m2 both connected to the cylinder then the system is at rest
so we can use torque balance here
Part b)
When block m_2 is removed then system becomes unstable
so force equation of mass m1
also we have
now we have
so angular acceleration is given as
Part c)
Tension in the rope is given as
Answer:
6.57 m/s
Explanation:
First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement
F=kx; F=180(.3) = 54 N
Next from Newton's second law find the acceleration of the mass.
Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²
Now use the kinematic equation for velocity (or speed)
v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.
v₀=0, since the mass is at rest before we release it
a=72 m/s² (from above)
x₀=0 as the start position already compressed
x₂=0.3m (this puts the spring back to it's natural length)
v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²
v₂= = 6.57 m/s