Calculating Tension and Speed in Circular Motion: A Tether Ball Example

AI Thread Summary
To calculate the tension in the cord and the speed of the tether ball, it's important to recognize that the ball moves in a horizontal circular path while the cord makes an angle of 20 degrees with the vertical. The tension in the cord provides the necessary centripetal force to maintain this circular motion. The tension and speed remain constant throughout the motion. Understanding the relationship between the angle, gravitational force, and centripetal force is crucial for solving the problem. Accurate calculations will yield the required values for tension and speed.
Lalasushi
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A tether ball of mass 0.5kg is attached to a vertical pole by a cord 1.5m long. Assume the cord attaches to the centre of the ball. If the cord makes an angle of 20 degrees with the vertical, then:
a) what is the tension in the cord?
b) what is the speed of the ball?

im having trouble with this question. I am assuming that the ball is nearly at the top of its swing...just 20 degrees away from it...can anyone help me with part a)? I am not sure if it has anything to do with centripetal force...
 
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You have misunderstood the situation:
The tether ball moves in a circular orbit in the HORIZONTAL plane, to which the the vertical is the normal direction.

In this situation, you will find that the speed of the ball and the tension in the cord are CONSTANTS throughout the circular motion.
 
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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