Answered

Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest at the top of an incline plane. The height of top of the incline plane is 1.2 m. Find translational speed of each cylinder upon reaching the bottom and determine which cylinder has the greatest translational speed upon reaching the bottom. Moment of inertia of hollow cylinder about its axis passing through the center is mr2 and for solid cylinder mr2/2

Sagot :

oladayoademosu

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.