Math Problem Statement

A sphere of mass m is attached to a string of length l. The sphere swings in a horizontal circle with a constant speed, making an angle θ with the vertical. Draw the forces acting on the sphere and derive an expression for its speed.

Solution

To assist you with the physics question in the image, I will provide explanations and solutions for the following parts:

Part (a)

i. Drawing and labeling the forces

The forces acting on the sphere are:

  1. Tension (TT): Acts along the string and points upward toward the pivot point.
  2. Gravitational force (mgmg): Acts vertically downward due to the sphere's weight.
  3. Centripetal force (FcF_c): Acts horizontally toward the center of the circular path.

These forces are not directly drawn on the string; rather, the components of TT resolve into two parts:

  • Ty=TcosθT_y = T \cos \theta: Balances mgmg vertically.
  • Tx=TsinθT_x = T \sin \theta: Provides the centripetal force.

When drawing:

  • Use a dot to represent the sphere.
  • Draw vectors TT, mgmg, and FcF_c with appropriate relative magnitudes based on known variables.

ii. Derivation for the speed of the sphere

Using Newton's second law:

  1. In the vertical direction: Tcosθ=mgT \cos \theta = mg Solve for TT: T=mgcosθT = \frac{mg}{\cos \theta}

  2. In the horizontal direction: Tsinθ=Fc=mv2rT \sin \theta = F_c = \frac{mv^2}{r} Substitute TT from the first equation: mgcosθsinθ=mv2r\frac{mg}{\cos \theta} \sin \theta = \frac{mv^2}{r}

  3. Simplify to solve for v2v^2: v2=grtanθcosθv^2 = \frac{gr \tan \theta}{\cos \theta} Substituting r=sinθr = \ell \sin \theta (radius in terms of string length): v2=gtanθsinθv^2 = g \ell \tan \theta \sin \theta

    Finally: v=gtanθsinθv = \sqrt{g \ell \tan \theta \sin \theta}

Part (b)

To evaluate the claim about the string being horizontal:

  • If the string were perfectly horizontal (θ=90\theta = 90^\circ), the tension would only provide centripetal force, and no vertical component of tension would exist to counteract gravity.
  • This is physically impossible as gravity (mgmg) must always be balanced by a vertical component of force.

Conclusion

  • The student's claim is incorrect. The string cannot be perfectly horizontal due to the requirement of a vertical force component to counteract gravity.

If you'd like further clarifications or a detailed diagram to visualize the solution, let me know.


Here are five related questions to deepen your understanding:

  1. What would happen to the speed of the sphere if the angle θ\theta increases?
  2. How does the tension in the string change with the length of the string?
  3. How would the forces change if the sphere's mass doubled?
  4. What role does centripetal force play in uniform circular motion?
  5. Why can't a perfectly horizontal string support the sphere's motion?

Tip: Always resolve forces into components when dealing with inclined or circular motion problems. This simplifies complex interactions into manageable equations.

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Math Problem Analysis

Mathematical Concepts

Circular Motion
Newton's Second Law
Force Resolution

Formulas

T * cos(θ) = mg
T * sin(θ) = mv²/r
v = √(g * l * tan(θ) * sin(θ))

Theorems

Newton's Second Law
Centripetal Force Equation

Suitable Grade Level

Grades 11-12