Chapter 6: Circular Motion and Other Applications of Newton's Laws

6.1 Extending the Particle in Uniform Circular Motion Model

A force causing a centripetal acceleration acts toward the center of the circular path.
This force causes a change in the direction of the velocity vector.

A small ball of mass m is suspended from a string of length L.
The ball revolves with constant speed v in a horizontal circle of radius r.
The system is known as a conical pendulum because the string sweeps out the surface of a cone.
Objective: Find an expression for v in terms of the geometry.
Forces in the vertical direction: \Sigma F_y = T \cos \theta - mg = 0
- Equation 1: T \cos \theta = mg
Forces in the horizontal direction:
\Sigma F_x = T \sin \theta = ma = m \frac{v^2}{r}
Solving for v:
- \tan \theta = \frac{v^2}{rg}
- v = \sqrt{rg \tan \theta}
Using geometry: r = L \sin \theta
- v = \sqrt{Lg \sin \theta \tan \theta}
Approximation for small angles:
- When \theta < 10^\circ,
  \sin \theta \approx \theta
- \tan \theta \approx \theta
- v \approx \sqrt{Lg \theta^2} = \theta \sqrt{Lg}

A puck of mass 0.500 kg is attached to a cord 1.50 m long.
The puck moves in a horizontal circle.
The cord can withstand a maximum tension of 50.0 N.
Objective: Find the maximum speed at which the puck can move before the cord breaks.
Assume the string remains horizontal during the motion.
The tension in the cord provides the centripetal force:
- T = m \frac{v^2}{r}
Solving for v:
- v = \sqrt{\frac{Tr}{m}}
Maximum speed:
- v{max} = \sqrt{\frac{T{max}r}{m}} = \sqrt{\frac{(50.0 \text{ N})(1.50 \text{ m})}{0.500 \text{ kg}}} = 12.2 \text{ m/s}

A 1500-kg car moves on a flat, horizontal road and negotiates a curve.
The radius of the curve is 35.0 m.
The coefficient of static friction between the tires and dry pavement is 0.523.
Objective: Find the maximum speed the car can have and still make the turn successfully.
The maximum static friction provides the centripetal force:
- f{s,max} = m \frac{v{max}^2}{r}
Vertical forces:
- \Sigma F_y = n - mg = 0
- n = mg
Maximum static friction:
- f{s,max} = \mus n = \mu_s mg
Setting the maximum static friction equal to the centripetal force:
- \mus mg = m \frac{v{max}^2}{r}
Solving for v_{max}:
- v{max} = \sqrt{\mus gr} = \sqrt{(0.523)(9.80 \text{ m/s}^2)(35.0 \text{ m})} = 13.4 \text{ m/s}

The car travels the curve on a wet day and begins to skid when its speed reaches only 8.00 m/s.
What can we say about the coefficient of static friction in this case?

A child of mass m rides on a Ferris wheel.
The child moves in a vertical circle of radius 10.0 m at a constant speed of 3.00 m/s.
Objective (A): Determine the force exerted by the seat on the child at the bottom of the ride.
- Express the answer in terms of the weight of the child, mg.
Free-body diagram at the bottom:
- Normal force n_{bot} is upward.
- Weight mg is downward.
Net force at the bottom:
- \Sigma F = n_{bot} - mg = m \frac{v^2}{r}
Normal force at the bottom:
- n_{bot} = mg + m \frac{v^2}{r} = mg \left(1 + \frac{v^2}{rg}\right)
- n_{bot} = mg \left(1 + \frac{(3.00 \text{ m/s})^2}{(10.0 \text{ m})(9.80 \text{ m/s}^2)}\right) = 1.09mg
Objective (B): Determine the force exerted by the seat on the child at the top of the ride.
- Free-body diagram at the top:
  - Normal force n_{top} is downward.
  - Weight mg is downward.
- ΣF = mg + n_top = m\frac{v^2}{r}
- Normal force at the top:
  - n_top= m\frac{v^2}{r} - mg
  - n_{top} = m \frac{v^2}{r} - mg = mg \left(\frac{v^2}{rg} - 1\right) = mg \left( \frac{(3.00 \text{ m/s})^2}{(10.0 \text{ m})(9.80 \text{ m/s}^2)} - 1\right) = 0.908mg

When a particle moves with varying speed in a circular path, there is, in addition to the radial component of acceleration, a tangential component having magnitude |dv/dt|.

A bead slides at constant speed along a curved wire lying on a horizontal surface.
- Draw the vector representing the force exerted by the wire on the bead at points A, B, and C.
Suppose the bead speeds up with constant tangential acceleration as it moves toward the right.
- Draw the vectors representing the force on the bead at points A, B, and C.

A small sphere of mass m is attached to the end of a cord of length R and set into motion in a vertical circle about a fixed point O.
Objective: Determine the tangential acceleration of the sphere and the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle \theta with the vertical.
Tangential acceleration:
- Forces along the x-axis: m g \sin \theta = m a_t
- a_t = g \sin \theta
Tension in the cord:
- Forces along the y-axis:
  \Sigma F_y = T - mg \cos \theta = \frac{mv^2}{R}
- T = mg \cos \theta + \frac{mv^2}{R}
- T = mg \left(\frac{v^2}{Rg} + \cos \theta\right)

T = mg \left(\frac{v^2}{Rg} + \cos \theta\right)
1. When \theta = 90^\circ or 270^\circ, \cos 90^\circ = 0, then: T = m \frac{v^2}{R}
  - The centripetal force just causes its acceleration is a_t = g \sin \theta = g \times 1 = g
2. At the top,
  \theta = 180^\circ, \cos(180^\circ) = -1
  T = mg \left(\frac{v^2}{Rg} - 1\right)
3. At the bottom,
  \theta = 0^\circ, \cos(0^\circ) = 1
  T_{bottom} = mg \left(\frac{v^2}{Rg} + 1\right)
Note: These equations are related to those in Example 6.5:
- n_{top} = mg \left(1 - \frac{v^2}{rg}\right)
- n_{bottom} = mg \left(1 + \frac{v^2}{rg}\right)
Keep in mind that:
- n is always upward and v is constant in Example 6.5.
- In this example,
  \theta changes according to position, and T is always toward the center.