Will the radial acceleration be constant?

AI Thread Summary
When torque is applied to a body in uniform circular motion, the radial acceleration is not constant due to changes in angular momentum. The application of torque increases angular momentum, leading to variations in angular velocity and subsequently altering radial acceleration. The discussion highlights two scenarios for changes in angular momentum: changes in magnitude and changes in rotation. While the initial argument addresses the first scenario, it raises the question of how the second scenario affects radial acceleration. Understanding these dynamics is crucial for analyzing motion under applied torque.
Amith2006
Messages
416
Reaction score
2
Sir,
When a torque is applied to a body undergoing uniform circular motion, will the radial acceleration be constant?
 
Physics news on Phys.org
Torque is defined as the time derivative of angular momentum. What do you think?
 
I think that when a torque is applied, the angular momentum increases as a result of which there a change in angular velocity and hence the radial acceleration changes. Is it right?
 
Sounds good so far. There are two ways that the angular momentum could change though, it could change its magnitude or it could rotate. You've argued what happens in the first case, but what about the second?
 
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...
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}...
Back
Top