Calculating Centripetal Acceleration & Tension in Circular Motion

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To calculate the centripetal acceleration of a 6.50-kg hammer being whirled in a horizontal circle with a radius of 2.0 m at a rate of 1.1 revolutions per second, the velocity must first be converted to meters per second. The formula for centripetal acceleration is a = (v^2)/r, where v is the linear velocity. The linear velocity can be determined by converting the rotational speed: v = (1.1 rev/s)(2π(2.0 m)), resulting in the necessary value for use in the acceleration formula. Additionally, the tension in the chain can be calculated by considering both the centripetal force and the weight of the hammer. Understanding these calculations is essential for accurately determining the dynamics of circular motion.
badboyben03
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An athlete whirls a 6.50-kg hammer tied to the end of a 2.0-m chain in a horizontal circle. The hammer moves at the rate of 1.1 rev/s.

What is the centripetal acceleration of the hammer?
What is the tension in the chain?
 
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What have you tried so far?

a = (v^2)/r is a start
 
1.1 rev/s is not the velocity right?
 
It is, but you must convert to m/s for use in the equation for centripetal acceleration.
 
how do you convert it?
 
(1.1 rev/s)(2piR meters/rev)
 
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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