Question

A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of 5 m/s. a. How far will the cylinder go up the plane?b. How long will it take to return to the bottom?

Answer 1

Hii dear,

◆ Answer-
s = 3.75 m
t = 1.533 s

◆ Explaination-
# Given-
v = 5 m/s
θ = 30°

# Solution-
When the cylinder go up the plane upto a height h, K.E. is converted into P.E.

By law of conservation of energy-
KEi + PEi = KEf + PEf
1/2 mv^2 + 1/2 Iω^2 + 0 = 0 + mgh
1/2 mv^2 + 1/2 (1/2 mr^2)ω^2 = mgh
1/2 mv^2 + 1/4 mv^2 = mgh
3/4 mv^2 = mgh
h = 3/4 v^2/g
h = 3/4×25/10
h = 1.875 m

Let s be the distance covered, then as
sinθ = h/s
s = h/sinθ
s = 1.875/sin 30°
s = 1.875/0.5
s = 3.75 m

Time taken to return to the bottom-
t = √[2s(1+k^2/r^2)]/√[gsinθ]
t = √[2×3.75(1+1/2)]/√[10/2]
t = √(11.75/5)
t = 1.533 s

Hope that was useful...