A force F required to keep a 5 kg mass moving round a cycle of radius 3.5 m at a speed

Question

A force F required to keep a 5 kg mass moving round a cycle of radius 3.5 m at a speed of 7 ms^-1. What is the speed, if the force is tripled?

The final answer to this question (the speed, when we triple the force) is 12.1 m/s

Data: Unrevealed information from the question

Mass of the object, m = 5 kg

Radius of the circle, r = 3.5 m

Velocity of the object, V = 7 m/s

Unknown: Unrevealed information from the question

Force towards the center (centripetal force), F = ?

When we triple the force, F_x3 = ?

Speed when we triple the force, V_x3 = ?

Formula: The equation that will help us solve the problem

Step 1: To find the force towards the center, we will apply the formula F = mV² / r

We will then multiply the result from the value of centripetal force by 3 to obtain F_x3

Step 2: We will use the formula F_x3 = mV_x3² / r to make the speed subject of the formula

V_x3 = √(F_x3 r / m) [We will apply this formula to find the speed after tripling our force]

Step 1: We will insert our data into the formula F = mV² / r

F = mV² / r = (5 x 7²) / 3.5 = 70 N

After tripling the force, F_x3 = 3 x 70 = 210 N

Step 2: Similarly, we will equally apply the formula V_x3 = √(F_x3 r / m) to find the speed

V_x3 = √(F_x3 r / m) = √(210 x 3.5 / 5) = √147

Hence,

The speed V_x3 = 12.1 m/s

Therefore, after tripling the force. The speed will become 12.1 meters per second.

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