It is trajectory acceleration. A friction track is a device to study motion in low friction environments, I believe. Does this help?
Answer:
the electromagnetic force
Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.
The pebble in the tread goes by 3 times every second.
This is the same as 3 times per second.
Recall the unit of frequency is Hertz or per second, s⁻¹
So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz
For angular motion:
Angular speed, ω = 2πf
= 2*π*3
= 6π rad/s
Linear speed, v = ωr = 6π * 0.3 = 1.8π m/s
Linear acceleration, a = v² / r
a = 1.8π * 1.8π / 0.3 = 10.8π² m/s²
Angular acceleration α = a/r = 10.8π² / 0.3 = 36π² rad/s²
Angular speed = 6π rad/s ≈ 18.840 rad/s
The linear speed of the pebble = 1.8π m/s ≈ 5.655 m/s
The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²
The linear acceleration of the pebble = 10.8π² m/s ≈ 106.592 m/s²
By definition, average acceleration will be:
a=(v(f)-v(i))/t
a=(2-14)/6=-2 (m/s^2)
________
-2(m/s^2)
