Physics: Circular Motion and Centripetal Acceleration
Calculating speed, period, force, and inward acceleration
Physics: Circular Motion and Centripetal Acceleration
Calculating speed, period, force, and inward acceleration
Physics - Grade 9-12
- 1
A car travels at a constant speed around a circular track. Explain why the car is accelerating even though its speed is not changing.
The car is accelerating because its velocity is changing direction. In circular motion, centripetal acceleration points toward the center of the circle even when speed stays constant. - 2
A ball moves in a circle with radius 0.75 m at a speed of 4.0 m/s. Calculate its centripetal acceleration.
Use a = v^2/r.
The centripetal acceleration is a = v^2/r = 4.0^2/0.75 = 21.3 m/s^2, so the ball accelerates toward the center at about 21 m/s^2. - 3
A satellite moves in a nearly circular orbit with radius 6.8 x 10^6 m and period 5.7 x 10^3 s. Calculate the satellite's orbital speed.
One full orbit covers a distance equal to the circumference of the circle.
The orbital speed is v = 2πr/T = 2π(6.8 x 10^6)/(5.7 x 10^3) = 7.5 x 10^3 m/s, so the satellite travels at about 7500 m/s. - 4
A 900 kg car rounds a flat curve of radius 50 m at 20 m/s. Calculate the centripetal force required to keep the car moving in the curve.
The required centripetal force is F = mv^2/r = 900(20^2)/50 = 7200 N. The force must point toward the center of the curve. - 5
A turntable rotates at 2.0 revolutions per second. A small object sits 0.40 m from the center. Find the period, speed, and centripetal acceleration of the object.
Frequency and period are related by T = 1/f.
The period is T = 1/f = 1/2.0 = 0.50 s. The speed is v = 2πr/T = 2π(0.40)/0.50 = 5.0 m/s. The centripetal acceleration is a = v^2/r = 5.0^2/0.40 = 63 m/s^2. - 6
At the top point of a circle, an object is moving clockwise. State the direction of its velocity and the direction of its centripetal acceleration at that instant.
At the top point, the velocity is tangent to the circle and points to the right for clockwise motion. The centripetal acceleration points downward toward the center of the circle. - 7
A child rides on a merry-go-round 2.5 m from the center. The merry-go-round completes one rotation every 8.0 s. Calculate the child's speed and centripetal acceleration.
The speed is v = 2πr/T = 2π(2.5)/8.0 = 2.0 m/s. The centripetal acceleration is a = v^2/r = 2.0^2/2.5 = 1.6 m/s^2 toward the center. - 8
A car rounds a flat curve of radius 80 m at 25 m/s. What minimum coefficient of static friction is needed to prevent skidding? Use g = 9.8 m/s^2.
On a flat curve, friction provides the centripetal force, so μmg = mv^2/r.
The minimum coefficient is μ = v^2/(rg) = 25^2/(80 x 9.8) = 0.80. The road needs a coefficient of static friction of at least 0.80. - 9
An object moves in a circle of fixed radius. If its speed doubles, how does its centripetal acceleration change?
Centripetal acceleration depends on the square of the speed.
The centripetal acceleration becomes four times larger because a = v^2/r. Doubling speed makes v^2 become 4 times as large. - 10
Two riders are on the same rotating platform. Rider A is 1.0 m from the center, and Rider B is 3.0 m from the center. The platform rotates once every 6.0 s. Which rider has the greater speed, and how many times greater is it?
Rider B has the greater speed. Since both riders have the same period, v = 2πr/T, so speed is proportional to radius. Rider B's speed is 3 times Rider A's speed. - 11
A 0.20 kg ball is swung in a vertical circle. At the bottom of the circle, the ball's speed is 6.0 m/s and the radius is 0.90 m. Find the net centripetal force required at the bottom.
The required net centripetal force is F = mv^2/r = 0.20(6.0^2)/0.90 = 8.0 N. At the bottom, the net force must point upward toward the center. - 12
Show that the formula a = v^2/r gives units of acceleration when v is measured in m/s and r is measured in m.
The units are (m/s)^2 divided by m, which equals m^2/s^2 divided by m. This simplifies to m/s^2, the correct unit for acceleration. - 13
A drone flies in a circular path at 12 m/s. Its centripetal acceleration is 3.0 m/s^2. Calculate the radius of the circular path.
Rearrange a = v^2/r to solve for r.
The radius is r = v^2/a = 12^2/3.0 = 48 m. The drone's circular path has a radius of 48 m. - 14
A student measures the speed of a cart moving in circles of different radii. For one trial, the radius is 0.50 m and the speed is 1.5 m/s. Calculate the centripetal acceleration. Then state whether increasing speed or increasing radius would increase the acceleration.
The centripetal acceleration is a = v^2/r = 1.5^2/0.50 = 4.5 m/s^2. Increasing speed would increase the acceleration, while increasing radius would decrease the acceleration if speed stayed the same. - 15
A road curve is banked so that a car can travel around it without friction. The curve radius is 60 m and the design speed is 18 m/s. Calculate the banking angle. Use tan θ = v^2/(rg) and g = 9.8 m/s^2.
After finding tan θ, use the inverse tangent function to find the angle.
The tangent of the angle is tan θ = 18^2/(60 x 9.8) = 0.551. Therefore θ = tan^-1(0.551) = 29 degrees, so the curve should be banked at about 29 degrees.