# Answer to Question #3694 in Mechanics | Relativity for Syed Salik Ali

Question #3694

Q-1 A uniform solid sphere of mass M and radius R hangs from a string of length R/2. Suppose the sphere is released from an initial position making an angle of 45 with the vertical
a) Calculate the angular velocity of the sphere when it swings through the vertical position.
b) Calculate the tension in the string at this instant.
Q-2 Two automobiles both of 1200kg and both trvelling at 30km/h collide on a frictionless icy road.They were initially moving on parallel paths in opposite directions, with a center -to -center distance of 1.0m. In the collision, the moment of inertia of this body about its center of mass is 2.5*10 power 3 kgm power 2.
a)Calculate the angular velocity of yhe wreck.
b) Calculate the kinetic energy before and after the collision. What is the change in Kinetic Energy?

Expert's answer

a) From the energy conservation law we get :

(mv

Where h is the difference of heights. Angular velocity is:

ω = v/r

Where r is the radius of a circle. In our case:

r = R/2

So

(mv

v

v

(ω

h=R/2 - R/2×cos45° = R/2(1-√2/2)

ω=√(8gh/R)=√(8g(R/2 (1-√2/2))/R^2) = √(4g(1-√2/2)/R)

b) Tension is the force that makes ball move downwards:

T = (mv

From the a) we have v

hence

T = 2mgh/R+mg = mg(1+2h/R)

(mv

^{2})/2 = mghWhere h is the difference of heights. Angular velocity is:

ω = v/r

Where r is the radius of a circle. In our case:

r = R/2

So

(mv

^{2})/2=mghv

^{2}/2=ghv

^{2}=2gh(ω

^{2}R^{2})/4=2ghh=R/2 - R/2×cos45° = R/2(1-√2/2)

ω=√(8gh/R)=√(8g(R/2 (1-√2/2))/R^2) = √(4g(1-√2/2)/R)

b) Tension is the force that makes ball move downwards:

T = (mv

^{2})/2+mgFrom the a) we have v

^{2}= 2ghhence

T = 2mgh/R+mg = mg(1+2h/R)

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