Tension and Speed Decrease: Determining Tension for Ball at θ = 20°

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Homework Help Overview

The problem involves a ball with a mass of 30 kg and an initial speed of 4 m/s at the lowest point of its motion, specifically at an angle θ = 0°. The task is to determine the tension in the cord and the rate at which the ball's speed decreases when the angle is θ = 20°. The context includes concepts from dynamics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of tension and the velocity of the ball at different angles, questioning the assumption that the speed remains constant from θ = 0° to θ = 20°.

Discussion Status

Some participants have provided guidance on using energy conservation to find the velocity at θ = 20°, suggesting that the velocity will differ from the initial speed. There is an acknowledgment of differing answers compared to a textbook, indicating ongoing exploration of the problem.

Contextual Notes

Participants note that the ball is not in uniform circular motion and that the tension must be evaluated at the specific angle of θ = 20°, which introduces complexities in the calculations.

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Homework Statement



The Ball has a mass of 30 kg and a speed v = 4 m/s at the instant it is at its lowest point, θ = 0°. Determine the tension in the cord and the rate at which the ball's speed is decreasing at the instant θ = 20°. Neglect the size of the ball.

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Homework Equations



ΣF = ma
an = v2

The Attempt at a Solution



T = mg + mv2
= 30kg(9.81 m/s2) + 30kg(4 m/s2)
= 414.3 N
 
Last edited:
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That looks correct. So onto the second part.
 
I would check the velocity at the point θ = 20º, because as I'm seeing in your attempt you're using the same very velocity as θ = 0º.
 
rock.freak667 said:
That looks correct. So onto the second part.

It doesn't agree with my textbook's answer, T = 361 N
 
red123 said:
It doesn't agree with my textbook's answer, T = 361 N

If that was the case, then v≠4 m/s.
 
Well, yeah the thing is that v = 4m/s at θ = 0º; but it is clear that at θ = 20º the velocity should differ, which should be easily calculated by energy conservation.
 
Redsummers said:
Well, yeah the thing is that v = 4m/s at θ = 0º; but it is clear that at θ = 20º the velocity should differ, which should be easily calculated by energy conservation.

If you use the tension given as the answer and work towards a velocity, it will be less than 4.
 
Yeah that's how it should be. For example:

E_t = mgh + 1/2mv^2 so, as the h increases according to the angle, the potential will also do so, while the velocity will drop.
 
red123 said:

Homework Equations



an = v2

Your problem occurred right here. Note that the ball is not moving in uniform speed, it drops down before it completes a loop. Therefore, it is not uniform circular motion and this equation does not work over here. Also note that naturally, tension force does not exceed the bigger external force on the bodies. That means at the lowest point, the tension equals to the force of gravity.
Another way of thinking of it is optimization. A graph reaches it's maximum when the slope of the tangent is 0. In a v-t graph, the slopes of tangents are the accelerations of the specific instants. Since the slope of the tangent is 0 at the lowest point, therefore the acceleration is 0. This leads to F_net=0.
 
  • #10
Try using conservation of energy.

red123 said:
Determine the tension in the cord and the rate at which the ball's speed is decreasing at the instant θ = 20°. Neglect the size of the ball.

Another problem you had was this. you are to find tension and acceleration at 20o
 

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