Finding Acceleration in Circular Kinematics without Degrees or Radians

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To find acceleration in circular kinematics, the centripetal acceleration formula, a = v²/r, is applied, where v is the velocity and r is the radius. Given a radius of 0.27m and a velocity of 5.09m/s, the calculated acceleration is 96.0 m/s². This value is considered large due to the wheel completing three revolutions per second, indicating a high rotational speed. Understanding the context of rapid rotation helps clarify the significance of the calculated acceleration. The discussion emphasizes the relationship between velocity, radius, and acceleration in circular motion.
rottweiler
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1. I have a circular kinematics question that I am not sure how to solve part of. I have a radius of.27m, and the wheel rotates three full circles a second. The first part of the question asked me to solve for velocity and I got 5.09m/s. I have to find acceleration in m/s^2. I am not allowed to use degrees only radians.



2. The four kinematics equations rearranged for polar coordinates.



3. Part one was ((2*Pi*3*27m)=5.09m/s I am not sure how to proceed with the acceleration[/b
 
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Do you know the equation governing centripetal acceleration?
 
(V^2)/r but that would make the acceleration 96.0 m/s^2 isn't that a large acceleration?
 
Last edited:
rottweiler said:
(V^2)/r but that would make the acceleration 96.0 m/s^2 isn't that a large acceleration?

yes that is what it would be, but you must realize that it is making 3 revolutions in 1 second. which is quite fast.
 
Thank you so much!
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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