Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
90 views
in Physics by (67.7k points)
closed by
A wheel of radius 10 cm can rotate freely about its centre as shown in figure. A stirng is wrapped over its rim and is ulle dby a force of 5.0 N. It is found that the torque roduces an angular acceleration 2.0 `rad/s^2` in the wheel. Calculate the moment of inertia of the wheel.
image

1 Answer

0 votes
by (75.2k points)
selected by
 
Best answer
The forces acting on the wheel are i. W due to gravity ii N due to the support at the cnetre and iii F due to tension. The torque of W and N are separately zero and that of F is F.r. The net torque is
`Gamma=(5.0N).(10cm)=0.50N.m.`
The moment of inertia is
`I =Gamma/a=(0.50N-m)/(2rads62)=0.25kg-m^2`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...