Calculate Angular Acceleration of Rotational Motion: 4.9rev in 1s

AI Thread Summary
To calculate the angular acceleration of a tire that completes 4.9 revolutions in 1 second, the correct approach involves recognizing that the motion is not at constant angular velocity. The conversion from revolutions to radians yields an angular speed of 30.7876 rad/s. However, since angular velocity changes over time, the formula a = w/t is inappropriate. Instead, a kinematic equation analogous to s = (1/2)at^2 should be used to find the angular acceleration. This clarification helps in correctly solving the problem.
islanderman7
Messages
5
Reaction score
0

Homework Statement


A tire placed on a balancing machine in a service station starts from rest and turns through 4.9 revolutions in 1.0 s before reaching its final angular speed. Calculate its angular acceleration.

I need my answer in
rad/s^2


Homework Equations


a=w/t
1rev= 2n rad


The Attempt at a Solution


Converted rev/s to rad/s
4.9rev/s * (2n rad/s)/ 1 rev= 30.7876 rad/s

a=w/t
a= (30.7876 rad/s)/ 1sec => 30.7876 rad/s^2

This answer comes out wrong.

What am I doing wrong?
 
Physics news on Phys.org
It's not a constant angular velocity question. Angular velocity changes with time, so you can't just use a=w/t. You need a formula analogous to s=(1/2)*a*t^2 for linear motion. It looks very similar, just the meaning of the letters changes.
 
Okay thanks. I get it now.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top