Calculating Speed at Top of Vertical Circle

AI Thread Summary
To determine the minimum speed a ball must have at the top of a vertical circle to keep the cord taut, the relevant formula involves the gravitational force and centripetal force. The calculation begins with the equation V min = √(gR), where g is the acceleration due to gravity and R is the radius of the circle. After substituting the values, the correct minimum speed is calculated as approximately 2.16 m/s, not 8.63 m/s as initially suggested. It's crucial to understand that at the top of the circle, the tension in the cord is zero, meaning only gravity provides the necessary centripetal force. This analysis emphasizes the importance of correctly applying physics principles to solve the problem.
kimikims
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I'm not sure what formula to use on this?

A ball of mass 15.9 g is attached to a cord of
length 0.478 m and rotates in a vertical circle.
The acceleration of gravity is 9.8 m/s^2

What is the minimum speed the ball must
have at the top of the circle so the cord does
not become slack? Answer in units of m/s.
 
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For all your questions, please show us what you have tried first. Begin with a free-body diagram of the ball at the top of the circle.
 
V min = Square root [(Ms)(g)(R)]

=Square root (15.9)(0.478)(9.8)

=Square root (74.48196)

=8.63 m/s ?
 
Now, remember...this is minumum speed...

...so the tension of the string is mimumum, or zero when the ball is at the top of its path (thus, only the force of gravity would equal the centripetal force.)

With this knowledge, draw a free body diagram of the ball when it is on the top of the path...
 
Last edited:
thermodynamicaldude said:
Now, remember...this is minumum speed...

...so the tension of the string is mimumum, or zero when the ball is at the top of its path (thus, only the force of gravity would equal the centripetal force.)

With this knowledge, draw a free body diagram of the ball when it is on the top of the path...


So would it be the square root of the radius times gravity?

= square root (0.478)(9.8)

= 2.16 m/s^2 ??
 
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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