A “swing” ride at a carnival consists of chairs that are swung in a circle by 15.0 m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 338 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair. Answer both parts to 3 significant figures. (a) Number 6620 N (b) Number 9.21 x m/s 15.0 m 60.0°
The Correct Answer and Explanation is:
Given Data:
- Length of cable, L=15.0 mL = 15.0 \, \text{m}L=15.0m
- Mass of chair and occupant, m=338 kgm = 338 \, \text{kg}m=338kg
- Angle of cable with vertical, θ=60.0∘\theta = 60.0^\circθ=60.0∘
- Gravitational acceleration, g=9.81 m/s2g = 9.81 \, \text{m/s}^2g=9.81m/s2
Part (a) Finding the Tension in the Cable
There are two forces to consider:
- Vertical component of tension: Balances the weight Tcos(θ)=mgT \cos(\theta) = mgTcos(θ)=mg Solving for TTT: T=mgcos(θ)T = \frac{mg}{\cos(\theta)}T=cos(θ)mg
Substituting values:T=338×9.81cos(60.0∘)=3316.980.5=6633.96 NT = \frac{338 \times 9.81}{\cos(60.0^\circ)} = \frac{3316.98}{0.5} = 6633.96 \, \text{N}T=cos(60.0∘)338×9.81=0.53316.98=6633.96N
Rounded to three significant figures:T=6630 NT = 6630 \, \text{N}T=6630N
Part (b) Finding the Speed of the Chair
The horizontal component of the tension provides the centripetal force:Tsin(θ)=mv2rT \sin(\theta) = \frac{mv^2}{r}Tsin(θ)=rmv2
Where:
- rrr is the radius of the circular path, calculated as:
r=Lsin(θ)=15.0×sin(60.0∘)=15.0×0.866=12.99 mr = L \sin(\theta) = 15.0 \times \sin(60.0^\circ) = 15.0 \times 0.866 = 12.99 \, \text{m}r=Lsin(θ)=15.0×sin(60.0∘)=15.0×0.866=12.99m
Solving for vvv:Tsin(θ)=mv2r ⟹ v2=Tsin(θ)rmT \sin(\theta) = \frac{mv^2}{r} \implies v^2 = \frac{T \sin(\theta) r}{m}Tsin(θ)=rmv2⟹v2=mTsin(θ)r
Substitute known values:v2=6633.96×sin(60.0∘)×12.99338=6633.96×0.866×12.99338v^2 = \frac{6633.96 \times \sin(60.0^\circ) \times 12.99}{338} = \frac{6633.96 \times 0.866 \times 12.99}{338}v2=3386633.96×sin(60.0∘)×12.99=3386633.96×0.866×12.99
Calculate step by step:6633.96×0.866=5744.896633.96 \times 0.866 = 5744.896633.96×0.866=5744.895744.89×12.99=74677.375744.89 \times 12.99 = 74677.375744.89×12.99=74677.3774677.37338=220.91\frac{74677.37}{338} = 220.9133874677.37=220.91v=220.91≈14.87 m/sv = \sqrt{220.91} \approx 14.87 \, \text{m/s}v=220.91≈14.87m/s
Final answer to three significant figures:v=14.9 m/sv = 14.9 \, \text{m/s}v=14.9m/s
Final Answers
(a) Tension in the cable: 6630 N6630 \, \text{N}6630N
(b) Speed of the chair: 14.9 m/s14.9 \, \text{m/s}14.9m/s
Explanation
The swing ride is a classic example of circular motion, where the tension in the cable provides both the vertical force to balance gravity and the horizontal force needed for circular motion. The problem requires careful resolution of forces into components, application of trigonometry, and correct algebra to find tension and speed. By using force equilibrium in the vertical direction and centripetal force in the horizontal direction, both unknowns are calculated accurately.
