Circular Motion & Gravitation — Drills · Grade 12 Science
Chapter: Forces, Rotation & EquilibriumWhat these drills covers
By the end of Circular Motion & Gravitation you should be able to:
- I can calculate centripetal acceleration (a = v²/r) and centripetal force (F = mv²/r) for uniform circular motion.
- I can apply Newton's law of universal gravitation (F = Gm₁m₂/r²) to find the gravitational force between two masses.
- I can explain how gravity supplies the centripetal force that holds a satellite in orbit.
6 questions, numeric, multiple choice and written answer, each with a worked explanation.
Vocabulary used: centripetal acceleration, centripetal force, period (T), Newton's law of universal gravitation, gravitational constant (G).
Problem 1
Answer and explanation
1.6 m/s²
a = v²/r = 8² ÷ 40 = 64 ÷ 40 = 1.6 m/s², directed toward the center.
Problem 2
Answer and explanation
4 N
F = mv²/r = 0.5 × 4² ÷ 2 = 0.5 × 16 ÷ 2 = 8 ÷ 2 = 4 N.
Problem 3
- forward, along the direction of motion
- backward, opposing the motion
- toward the center of the circle
- away from the center
Answer and explanation
toward the center of the circle
The net force in circular motion is centripetal — it points toward the center of the circle.
Problem 4
- unchanged
- twice as strong
- four times as strong
- half as strong
Answer and explanation
four times as strong
Force is proportional to m₁ × m₂. Doubling each mass multiplies the product by 2 × 2 = 4, so the force becomes four times as strong.
Problem 5
Answer and explanation
Earth's gravity supplies the centripetal force. The Moon moves nearly in a circle, so it needs a net force pointing toward Earth's center; gravitational attraction (F = Gm₁m₂/r²) provides exactly that inward pull, continually bending the Moon's path so it 'falls around' Earth instead of flying off in a straight line.
Problem 6
Answer and explanation
4000 N
F = mv²/r = 1,000 × 10² ÷ 25 = 1,000 × 100 ÷ 25 = 1,000 × 4 = 4,000 N.