# Centripetal force – problems and solutions

1. A 200-gram ball, attached to the end of a cord, is revolved in a horizontal circle with an angular speed of 5 rad s^{-1}. If cord’s length is 60 cm, what is the centripetal force?

__K____nown :__

Object’s mass (m) = 200 gr = 200/1000 kg = 2/10 kg = 0.2 kg

Angular speed (ω) = 5 rad/s

Cord’s length = radius (r) = 60 cm = 60/100 m = 0.6 m

__Wanted :__ The centripetal force

__Solution :__

The centripetal force is the resultant force that causes the centripetal acceleration.

The equation of the centripetal force :

∑F = m a_{s}

∑F = m v^{2}/r = m ω^{2} r

∑*F = **Centripetal force**, m = **object’s mass**, v = **linear velocity**, **ω = **angular velocity**, **r = **radius.*

∑F = m ω^{2} r = (0.2)(5)^{2 }(0.6) = (0.2)(25)(0.6) = 3 N

2. A stone attached at the end of a cord and rotated in a horizontal circle by a student. If the final speed of the stone = 2 x the initial speed, then what is the centripetal force.

__Known :__

Stone’s mass = m

Stone’s speed = v

Cord’s length = radius = r

__Wanted:__ The centripetal force

__Solution :__

3. A curve road of radius R is designed so that a car traveling at speed 10 ms^{–1 }can negotiate the turn safely. The coefficient of static friction between car and road = 0.5. What is the radius? Acceleration due to gravity (g) = 10 ms^{–2}.

__Known :__

Speed (v) = 10 m/s

The coefficient of static friction between car and road (μ_{s}) = 0.5

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted:__ Radius

__Solution :__

The only one force in the horizontal direction is the force of static friction. The equation of the static friction :

4. The coefficient of static friction between tire and road is 0.4. If acceleration due to gravity is 10 m/s^{2}, what is the maximum speed so the car can turn without skidding out of a curved path.

__Known :__

Coefficient of static friction (μ_{s}) = 0.4

Acceleration due to gravity (g) = 10 m/s^{2}

Radius of path (R) = 40 meters

__Wanted:__ maximum speed (v)

__Solution :__

**The equation of Newton’s second law in uniform circular motion :**

*ΣF = **centripetal force** = **net force**, m = m**ass**, a*_{s }*= **centripetal acceleration**, v = **linear speed**, R = **radius of path*

**Centripetal force**

Centripetal force is the net force which produces centripetal accelerations. In this case, the centripetal force is the force of static friction.

The equation of the force of static friction :

*μ*_{s}* = **coefficient of static friction**, w = **weight**, m = mass, g = **acceleration due to gravity*

**The maximum speed ****(v) :**