Physics Help

Question

7) An iron cylinder of mass 270 grams and radius 2.0 cm rolls (without slipping) down an incline. The vertical height of the incline is 44 cm and it makes an angle of 34° with the horizontal. Neglect air resistance in all parts of this problem.

a) If the initial velocity of the cylinder were zero, what would be the final speed of the cylinder at the base of the incline?

2.40 m/s.

b) What would its angular velocity, w , at the base of the incline be?

120 rad/sec.

 

 

Please show me how to get these!

Answer 1

(a)

 

The potential energy at the top converts fully to a combination of translational and rotational kinetic energy at the bottom.

 

mgh = (1/2)mv² + (1/2)Iω² 

 

I = (1/2)mR²

ω = v/R

 

mgh = (1/2)mv² + (1/2)[(1/2)mR²(v/R)²]

 

gh = v²/2 + (1/4)v² = (3/4)v²

 

v = √(4gh/3) = √(4*9.8*0.44/3) m/s = 2.398 m/s ≅ 2.40 m/s

 

(b)

 

ω = v/R = (2.40 m/s) /(0.02 m) = 120 rad/s