Worked examples · Sheet 3

Motions with conservation of mechanical energy

Theory used: Conservation of mechanical energy Mechanical energy Potential energy of the weight Kinetic energy Sign of the work

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A stone is thrown from the top of a tower of height $h = 15 m$ with a velocity of magnitude $v_{0} = 8.0 m/s$ . Compute the magnitude of the velocity with which it reaches the ground, neglecting air resistance, and determine whether it depends on the direction of the throw.

\frac{1}{2} m v_{0}^{2} + m g h = \frac{1}{2} m v^{2} + 0

v = v_{0}^{2} + 2 g h

v = 8. 0^{2} + 2 \cdot 9.81 \cdot 15 m/s = 64 + 294.3 m/s \approx 18.9 m/s

Neglecting air resistance, the only force acting on the stone is the weight, which is a conservative force, so by the conservation of mechanical energy the mechanical energy at the moment of the throw equals that on arrival at the ground. We choose zero height at the ground; at the start the energy is the sum of $\frac{1}{2} m v_{0}^{2}$ and $m g h$ , on arrival it is only kinetic. The mass cancels and we obtain $v \approx 18.9 m/s$ . Only the magnitude $v_{0}$ appears in the equation, and not the launch angle, therefore the magnitude of the arrival velocity does not depend on the direction of the throw; the trajectory, the time of flight and the landing point do depend on the direction. As a limiting case, with $v_{0} = 0$ we recover free fall, $v = 2 g h$ .

Exercise 2

A roller coaster car starts with negligible speed from the point $A$ , at height $20 m$ , and travels along a smooth track passing through the lowest point $B$ , at height $8.0 m$ , and then through the point $C$ , at height $14 m$ . Compute the speed of the car at $B$ and at $C$ .

y = 14 + 6 \cos \left(\dfrac{\pi x}{10}\right)

Profile of the track: the car starts from

A

at height

20 m

, passes through

B

at height

8.0 m

and through

C

at height

14 m

m g y_{A} = \frac{1}{2} m v_{B}^{2} + m g y_{B} ⟹ v_{B} = 2 g (y_{A} - y_{B})

v_{C} = 2 g (y_{A} - y_{C})

v_{B} = 2 \cdot 9.81 \cdot 12 m/s \approx 15.3 m/s, v_{C} = 2 \cdot 9.81 \cdot 6.0 m/s \approx 10.8 m/s

The track is smooth, so its reaction force is always perpendicular to the track and to the velocity, and it does no work. The only force that does work is the weight, therefore the mechanical energy is conserved. At $A$ the energy is only potential, because the speed is negligible; at $B$ and at $C$ it is the sum of kinetic and potential energy. The difference in height between $A$ and $B$ is $12 m$ and that between $A$ and $C$ is $6.0 m$ . The speed at a point depends only on the difference in height from the starting point, and not on the shape of the track between the two points. At $C$ the car is slower than at $B$ , because it has climbed $6.0 m$ , converting part of its kinetic energy into potential energy.

Exercise 3

A simple pendulum consists of a small ball of mass $m = 0.50 kg$ hanging from an inextensible string of negligible mass and length $ℓ = 1.2 m$ . The ball is released from rest with the string inclined at $θ_{0} = 6 0^{\circ}$ to the vertical. Compute the speed of the ball at the lowest point and the tension in the string at that point.

m g ℓ (1 - cos θ_{0}) = \frac{1}{2} m v^{2} ⟹ v^{2} = 2 g ℓ (1 - cos θ_{0})

T - m g = m \frac{v ^{2}}{ℓ} ⟹ T = m g + 2 m g (1 - cos θ_{0}) = m g (3 - 2 cos θ_{0})

v T = 2 \cdot 9.81 \cdot 1.2 \cdot (1 - 0.50) m/s = 11.77 m/s \approx 3.4 m/s = 0.50 \cdot 9.81 \cdot (3 - 1) N \approx 9.8 N

The tension of the string is directed along the string, while the velocity of the ball is tangent to the circle it describes, so the two are perpendicular and the tension does no work. Therefore the mechanical energy is conserved, with only the weight doing work. Between the initial position and the lowest point the ball drops by the height $ℓ - ℓ cos θ_{0}$ , because with the string inclined it is at a vertical distance $ℓ cos θ_{0}$ below the suspension point.

The tension cannot be obtained from energy, since it does no work, and it is found with Newton’s second law along the radial direction. At the lowest point the ball travels along a circle of radius $ℓ$ , so the resultant of tension and weight is the centripetal force $m \frac{v ^{2}}{ℓ}$ , pointing upward. Substituting $v^{2}$ obtained from energy we find $T = 2 m g \approx 9.8 N$ , twice the weight.

Exercise 4

A car of mass $m = 300 kg$ travels along a smooth track that ends in a vertical loop of radius $R = 4.0 m$ ; the car runs on the inside of the loop and can leave the track. Compute the minimum height $h$ , measured from the lowest point of the loop, from which the car must start from rest to go all the way around the loop, and the reaction force of the track at the lowest point in this case. The car is treated as a point particle.

The car starts from

A

and goes around the loop of radius

R

; at the highest point

C

, at the minimum speed, the only force is the weight, which provides the centripetal force

⎩ ⎨ ⎧ N_{C} + m g = m \frac{v _{C}^{2}}{R} N_{C} \geq 0 ⟹ v_{C}^{2} \geq g R

m g h = m g (2 R) + \frac{1}{2} m v_{C}^{2} ⟹ h = 2 R + \frac{v _{C}^{2}}{2 g} \geq 2 R + \frac{R}{2} = \frac{5}{2} R

N_{B} - m g = m \frac{v _{B}^{2}}{R}, v_{B}^{2} = 2 g h = 5 g R ⟹ N_{B} = 6 m g

h_{m i n} = \frac{5}{2} \cdot 4.0 m = 10 m, N_{B} = 6 \cdot 300 \cdot 9.81 N \approx 1.8 \cdot 1 0^{4} N

At the highest point $C$ the car is on the inside of the loop, so the reaction force of the track and the weight both point downward, that is, toward the center, and their sum is the centripetal force. The car stays in contact with the track as long as the reaction force is positive or zero; the two conditions must be satisfied together, and together they give $v_{C}^{2} \geq g R$ . The minimum speed corresponds to $N_{C} = 0$ , when the weight alone provides the centripetal force.

The track is smooth and its reaction force does no work, so the mechanical energy is conserved between the start and the point $C$ , which is at height $2 R$ . Substituting the condition on $v_{C}^{2}$ we obtain $h \geq \frac{5}{2} R$ , that is, a minimum height of $10 m$ . At the lowest point $B$ the car has dropped by $h$ , therefore $v_{B}^{2} = 2 g h = 5 g R$ ; here the reaction force points upward, toward the center, and the weight downward. From Newton’s second law along the radial direction we obtain $N_{B} = 6 m g$ , six times the weight, independently of the radius of the loop.