Circular Motion Practice Problems
Problem 1: Non-Uniform Circular Motion
A 1.5 kg object is attached to a 2 m-long string and whirled in a vertical circle with a variable speed.
a) Calculate the minimum speed the object must have at the top of the circle to just maintain tension in the string.
b) If the object’s speed at the bottom of the circle is 10 m/s, determine the tension in the string at this point.
c) Discuss how the tension changes as the object moves from the top to the bottom of the circle and explain why.
Problem 2: Conical Pendulum
A 0.8 kg object is tied to a 1.2 m string and swung in a horizontal circle, making a constant angle of 30° with the vertical.
a) Calculate the tension in the string.
b) Determine the object's speed in this conical motion.
c) Explain why the angle of the string with the vertical affects the tension and speed required to maintain the motion.
Problem 3: Banked Curve with Friction
A car with a mass of 1200 kg travels around a banked curve with a radius of 50 m. The curve is banked at an angle of 20°, and the coefficient of static friction between the tires and the road is 0.3.
a) Determine the minimum speed at which the car can travel around the curve without sliding down the bank.
b) Calculate the maximum speed the car can travel without skidding off the curve.
c) Discuss the role of friction in this scenario and explain how it contributes to keeping the car in circular motion.
Problem 4: Gravitational Force in Circular Orbits
A satellite of mass 200 kg is orbiting Earth in a circular path at an altitude where the gravitational field strength is 7.0 N/kg.
a) Calculate the gravitational force acting on the satellite.
b) Determine the satellite's orbital speed.
c) Explain why the gravitational force provides the necessary centripetal force for the satellite’s circular motion.
Problem 5: Rotational Speed and Acceleration in a Ferris Wheel
A Ferris wheel with a radius of 25 m rotates such that passengers experience a maximum apparent weight of 1.5 times their actual weight at the lowest point.
a) Calculate the rotational speed of the Ferris wheel.
b) Determine the normal force on a 60 kg passenger at the top and bottom of the Ferris wheel.
c) Explain how the apparent weight changes between the top and bottom positions and why this occurs in terms of circular motion concepts.
Problem 6: Centripetal Force on a Rotating Disc
A small 0.2 kg object rests on a disc that rotates at 1.5 revolutions per second. The object is located 0.4 m from the center of the disc.
a) Calculate the centripetal force required to keep the object in its circular path.
b) If the coefficient of static friction between the object and the disc is 0.4, determine if the object will stay in place.
c) Explain the role of static friction in providing the centripetal force and discuss what would happen if the rotation speed increased.
Problem 7: Satellite Motion and Gravitational Potential Energy
An 800 kg satellite is launched into a circular orbit around Earth at an altitude where the gravitational field strength is 5 N/kg.
a) Calculate the satellite's orbital radius from Earth's center.
b) Determine the gravitational potential energy of the satellite in this orbit.
c) Explain why a satellite in a stable orbit does not require additional energy to maintain its motion.
Problem 8: Artificial Gravity in a Rotating Space Station
A space station is designed as a rotating ring with a radius of 100 m, spinning to create artificial gravity.
a) Calculate the required tangential speed of the outer edge to simulate Earth’s gravity.
b) If a person stands at a distance of 80 m from the center, determine the artificial gravity they would experience.
c) Explain how changing the radius of rotation affects the perceived gravity within the station and discuss potential implications for station design.
Problem 9: Tension and Circular Motion in a Horizontal Swing
A child swings a 0.5 kg ball in a horizontal circle with a 1.5 m-long string. The ball completes each circle in 1.2 seconds.
a) Calculate the tension in the string during the swing.
b) Determine the centripetal acceleration of the ball.
c) Discuss the relationship between the speed of the ball and the tension in the string. Explain why increasing the speed requires a greater tension force.
Problem 10: Roller Coaster Loop-the-Loop
A roller coaster cart with a mass of 400 kg travels through a vertical loop of radius 10 m. At the top of the loop, the speed of the cart is 8 m/s.
a) Calculate the minimum speed needed at the top of the loop to prevent the cart from falling.
b) Determine the normal force acting on the cart at the top of the loop at 8 m/s.
c) Explain how energy conservation principles apply as the cart moves through the loop and why the cart’s speed changes between the top and bottom.
Answer Key for Circular Motion Practice Problems
Problem 1: Non-Uniform Circular Motion
a) Minimum speed at the top of the circle:
For just enough tension, T=0T = 0T=0 at the top, so mg=mv2rmg = \frac{mv^2}{r}mg=rmv2.
v=rg=2×9.8≈4.43 m/sv = \sqrt{rg} = \sqrt{2 \times 9.8} \approx 4.43 \, \text{m/s}v=rg=2×9.8≈4.43m/sb) Tension at the bottom:
T=mg+mv2r=1.5×9.8+1.5×1022≈86.7 NT = mg + \frac{mv^2}{r} = 1.5 \times 9.8 + \frac{1.5 \times 10^2}{2} \approx 86.7 \, \text{N}T=mg+rmv2=1.5×9.8+21.5×102≈86.7Nc) Explanation of tension change:
The tension is greatest at the bottom due to the combined effect of gravitational and centripetal forces. At the top, tension decreases because gravity contributes to the centripetal force.
Problem 2: Conical Pendulum
a) Tension in the string:
Tcos(30∘)=mg⇒T=mgcos(30∘)≈9.05 NT \cos(30^\circ) = mg \Rightarrow T = \frac{mg}{\cos(30^\circ)} \approx 9.05 \, \text{N}Tcos(30∘)=mg⇒T=cos(30∘)mg≈9.05Nb) Object’s speed:
Tsin(30∘)=mv2r⇒v=Tsin(30∘)⋅Lm≈2.57 m/sT \sin(30^\circ) = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{T \sin(30^\circ) \cdot L}{m}} \approx 2.57 \, \text{m/s}Tsin(30∘)=rmv2⇒v=mTsin(30∘)⋅L≈2.57m/sc) Effect of angle on tension and speed:
A larger angle increases the component of tension that provides the centripetal force, requiring a higher speed to maintain circular motion.
Problem 3: Banked Curve with Friction
a) Minimum speed without sliding:
With friction assisting gravity, vmin=grtan(20∘)×(1−μ)≈6.12 m/sv_{\text{min}} = \sqrt{g r \tan(20^\circ) \times (1 - \mu)} \approx 6.12 \, \text{m/s}vmin=grtan(20∘)×(1−μ)≈6.12m/sb) Maximum speed without skidding:
With friction opposing the outward force, vmax=gr(tan(20∘)+μ)≈12.4 m/sv_{\text{max}} = \sqrt{g r (\tan(20^\circ) + \mu)} \approx 12.4 \, \text{m/s}vmax=gr(tan(20∘)+μ)≈12.4m/sc) Role of friction:
Friction helps prevent sliding by providing additional force components parallel to the incline, aiding in keeping the car on the curve.
Problem 4: Gravitational Force in Circular Orbits
a) Gravitational force on the satellite:
F=mg=200×7=1400 NF = mg = 200 \times 7 = 1400 \, \text{N}F=mg=200×7=1400Nb) Orbital speed:
v=gr=7⋅R≈7.67 km/sv = \sqrt{gr} = \sqrt{7 \cdot R} \approx 7.67 \, \text{km/s}v=gr=7⋅R≈7.67km/sc) Explanation of gravitational force as centripetal force:
In orbit, gravity acts as the centripetal force required to maintain circular motion, balancing the satellite's inertia.
Problem 5: Rotational Speed and Acceleration in a Ferris Wheel
a) Rotational speed of the Ferris wheel:
mv2r=1.5×mg⇒v=1.5×25×9.8≈19.1 m/s\frac{mv^2}{r} = 1.5 \times mg \Rightarrow v = \sqrt{1.5 \times 25 \times 9.8} \approx 19.1 \, \text{m/s}rmv2=1.5×mg⇒v=1.5×25×9.8≈19.1m/sb) Normal force at the top and bottom:
At the bottom: Fnormal=1.5×60×9.8=882 NF_{\text{normal}} = 1.5 \times 60 \times 9.8 = 882 \, \text{N}Fnormal=1.5×60×9.8=882N
At the top: Fnormal=0.5×60×9.8=294 NF_{\text{normal}} = 0.5 \times 60 \times 9.8 = 294 \, \text{N}Fnormal=0.5×60×9.8=294Nc) Explanation of apparent weight changes:
Apparent weight increases at the bottom due to the centripetal force direction, while it decreases at the top because gravity assists the centripetal force.
Problem 6: Centripetal Force on a Rotating Disc
a) Centripetal force required:
Fc=mrω2=0.2×0.4×(2π×1.5)2≈7.11 NF_c = mr\omega^2 = 0.2 \times 0.4 \times (2 \pi \times 1.5)^2 \approx 7.11 \, \text{N}Fc=mrω2=0.2×0.4×(2π×1.5)2≈7.11Nb) Frictional force check:
fmax=μs⋅mg=0.4×0.2×9.8=0.784 Nf_{\text{max}} = \mu_s \cdot mg = 0.4 \times 0.2 \times 9.8 = 0.784 \, \text{N}fmax=μs⋅mg=0.4×0.2×9.8=0.784N.
Since fmax<Fcf_{\text{max}} < F_cfmax<Fc, the object will not stay in place.c) Effect of increased speed:
Increased speed increases the centripetal force, and without sufficient friction, the object will slide outward.
Problem 7: Satellite Motion and Gravitational Potential Energy
a) Orbital radius:
g=GMr2⇒r=GMg≈8.98×106 mg = \frac{GM}{r^2} \Rightarrow r = \sqrt{\frac{GM}{g}} \approx 8.98 \times 10^6 \, \text{m}g=r2GM⇒r=gGM≈8.98×106mb) Gravitational potential energy:
U=−GMmr=−3.57×109 JU = -\frac{GMm}{r} = -3.57 \times 10^9 \, \text{J}U=−rGMm=−3.57×109Jc) Stable orbit explanation:
In stable orbit, the satellite’s motion and gravitational force are balanced, requiring no additional energy input.
Problem 8: Artificial Gravity in a Rotating Space Station
a) Required tangential speed:
v=rg=100×9.8≈31.3 m/sv = \sqrt{rg} = \sqrt{100 \times 9.8} \approx 31.3 \, \text{m/s}v=rg=100×9.8≈31.3m/sb) Artificial gravity at 80 m:
gart=v2r=31.3280≈12.2 m/s2g_{\text{art}} = \frac{v^2}{r} = \frac{31.3^2}{80} \approx 12.2 \, \text{m/s}^2gart=rv2=8031.32≈12.2m/s2c) Effect of radius on perceived gravity:
Decreasing the radius increases perceived gravity, potentially affecting comfort and health of station occupants.
Problem 9: Tension and Circular Motion in a Horizontal Swing
a) Tension in the string:
T=mv2r=0.5×(2π×1.51.2)21.5≈16.4 NT = \frac{mv^2}{r} = \frac{0.5 \times (2 \pi \times \frac{1.5}{1.2})^2}{1.5} \approx 16.4 \, \text{N}T=rmv2=1.50.5×(2π×1.21.5)2≈16.4Nb) Centripetal acceleration:
ac=v2r=(2π×1.5/1.2)21.5≈32.8 m/s2a_c = \frac{v^2}{r} = \frac{(2 \pi \times 1.5 / 1.2)^2}{1.5} \approx 32.8 \, \text{m/s}^2ac=rv2=1.5(2π×1.5/1.2)2≈32.8m/s2c) Relationship between speed and tension:
Higher speeds increase the tension force due to the greater centripetal force required.
Problem 10: Roller Coaster Loop-the-Loop
a) Minimum speed at the top:
v=rg=10×9.8≈9.9 m/sv = \sqrt{rg} = \sqrt{10 \times 9.8} \approx 9.9 \, \text{m/s}v=rg=10×9.8≈9.9m/sb) Normal force at 8 m/s:
N=mv2r−mg=400×8210−400×9.8≈−1920 NN = \frac{mv^2}{r} - mg = \frac{400 \times 8^2}{10} - 400 \times 9.8 \approx -1920 \, \text{N}N=rmv2−mg=10400×82−400×9.8≈−1920N (indicating the cart would fall at this speed)c) Energy conservation in the loop:
As the cart moves through the loop, potential energy at the top converts to kinetic energy at the bottom, increasing its speed due to conservation of mechanical energy.