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
Messages
36
Reaction score
0
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.
 
Physics news on Phys.org
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 ??
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top