A stone tied to the end of a string of length 1m is whirled in a horizontal circle with a constant speed. If stone makes 22 revolution in 44second, what is the magnitude and direction of the acceleration of stone?
A
Acceleration will be π2m/s2 radially inwards.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Acceleration will be π2m/s2 radially outwards.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Acceleration will be π2m/s2 along the tangent to the circle.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Acceleration will be π4m/s2 radially inwards.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A Acceleration will be π2m/s2 radially inwards. Radial acceleration or centripetal accceleration is given by: ac=v2r=ω2r
Frequency of revolution is, f=2244=12s−1 ⇒ω=2πf=πrad/s ∴ac=ω2r=π2×1=π2m/s2
Tangential acceleration is the rate of change of speed w.r.t time: ⇒at=dvdt=0 ∵ speed is constant
Hence net acceleration of stone →a given by: →a=→ac+→at=→ac=π2m/s2 ∴|→a|=|→ac|=π2m/s2
Direction is along the radius towards the centre i.e. radially inwards.