Question

A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal. How long does it take it to travel the first 3.1 m?A) 1.4 sB) 1.1 sC) 2.1 sD) 1.6 sE) 1.8 s

Answer 1

Answer:

B ( 1.1 s )

Explanation:

Given that a thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal. 

Let the height = h

Sin 30 = h / 3.1

h = 3.1 Sin 30

The P.E at the top = mgh

P.E = 9.8 × 3.1 Sin 30 × m

P.E = 15.19m

The K.E = P.E

K.E = 1/2mv^2

1/2 × V^2 × m = 15.19m

The mass m will cancel out.

V^2 = 30.38

V = 5.51 m/s

Using third equation of motion to find acceleration

V^2 = U^2 + 2as

Since it starts from rest, initial velocity u = 0

30.38 = 2 × 3.1 × a

30.38 = 6.2a

a = 30.38 / 6.2

a = 4.9 m/s^2

Using equation one of linear motion to calculate time t

V = U + at

5.51 = 4.9t

t = 5.51 / 4.9

t = 1.12 s

Therefore, it will take approximately 1.1 s to travel the first 3.1 m

The correct option is B