Search Colleges, Courses, Exams, Questions and Articles

Ask a query on 8879100846 Ask Login | Sign Up

 

 

› Ask & Answer › Question

7.11 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.

0 8 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    Let m and r be the mass and radius of the hollow cylinder and solid sphere.

    The moment of inertia of the hollow cylinder about its standard axis, =I sin?36.9°)

    The MI of the solid sphere about an axis passing through its centre, Is = (2/5) mr2

    We know the relation cos?36.9° = I cos?53.1° , where

    α = angular acceleration

    τ = torque

    I = moment of inertia

    For the hollow cylinder, α = α

    For the solid sphere, τ = τh

    Since the torque applied is same, Iαh = τs , we get

    Isαs = τh = (mr2) /( (2/5) mr2 )) = 5/2

    Hence τs ……(i)

    Now using the relation

    αhαs + IsIh where

    αh<αs = ini

    ...more

Similar Questions for you

The position vector of 1kg object is r → = ( 3 i ^ − j ^ ) m and its velocity v → = ( 3 j ^ + k ^ ) m s − 1 . The magnitude of is angular momentum is x   N m  where x is________.
R
Raj Pandey

L ? = r ? * m v ?

= ( 3 i ^ ? j ^ ) * ( 3 j ^ + k ^ )

= 9 k ^ + 3 ( ? j ^ ) ? i ^

= ? i ^ ? 3 j ^ + 9 k ^

| L ? | = 1 + 9 + 8 1 = 9 1

Match List – I with List – II

Choose the correct answer from the options given below:
R
Raj Pandey

Use formula for M.I.

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1 . If the fragments having equal mass fly off along mutually perpendicular directions with speed v , the speed of the third (lighter) fragment is :
R
Raj Pandey

By conservation of momentum:

m ( 0 ) = 2 m 5 ( - v i ˆ ) + 2 m 5 ( - v j ˆ ) + m 5 v ? ' ⇒ v ? ' = 2 v i ˆ + 2 v j ˆ ⇒ v ' = ( 2 v ) 2 + ( 2 v ) 2 = 2 2 v

The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200rpm to 3120 in 16 seconds. The angular acceleration in r a d / s 2  is:
R
Raj Pandey

ω = ω 0 + α t α = ω - ω 0 t = ( 3120 - 1200 ) 16 s r p m = 1920 16 × 2 π 60 r a d / s 2 = 4 π r a d / s 2

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is :
R
Raj Pandey

k = 1 m

⇒ k 1 k 2 = I 1 I 2 = m R 2 / 2 m R 2 / 4 = 2 : 1

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Write a Review

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Ask Now
Community GuidelinesUser Points System
QuestionsDiscussionsTags
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post