wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A 40 kg child makes 10 revolutions in 29.3 seconds, moving on a circular path. The radius of the path taken by the child is 2.9 meters. Then, centripetal force acting on him is:

A
5.32 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
53.2 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
533.6 N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
267 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 533.6 N
Given, m=40 kg and
Radius of circular path(r) = 2.9 m
Time taken to complete 1 revolution
i.e T=29.310=2.93 s
T=2πωω=2πT

Now speed of boy
v=rω=2πrT=2×3.14×2.92.93=6.22 m/s

To find centripetal acceleration, applying the formula:
ac=v2r=(6.22)22.9=13.34 m/s2
Fcentripetal=m× ac
Fc=40×13.34=533.6 N

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon