A gramophone record takes 5s to reach its constant angular velocity of 4\(\pi\) rads^-1 from rest. Find its constant angular acceleration.

Derekchukwuemeka

3 years ago

From the given information, we can determine the angular acceleration of the gramophone record.
Angular acceleration (α) can be calculated using the following equation:
α = (ω - ω₀) / t
Where:
α = angular acceleration
ω = final angular velocity
ω₀ = initial angular velocity
t = time
Given:
ω = 4π rad/s
ω₀ = 0 rad/s
t = 5 s
Plugging in the values:
α = (4π - 0) / 5
= 4π / 5
= (4/5)π rad/s²
Therefore, the angular acceleration of the gramophone record is (4/5)π which is
approximately 0.8π rads^-2

oiza234

4 years ago

Just checked ALL INCLUSIVE CALCULATIONS IN PHYSICS (BY SOLOMON DAUDA YAKWO) and the answer was given as 20πrads–² as against the answer given as 0.8πrads–². Please explain ,I am confused.

couragechiemela

3 years ago

angular acceleration=angular velocity/time which is 4π /5=0.8πradsec-2