# 17.13 Centripetal force – throwing the hammer

Before you read this, I suggest you read posts 17.12, 17.11, 17.10 and 17.4. The man in the picture is practising “throwing the hammer”. In this event, the participant swings a spherical “hammer” using a cable, as shown in the picture, several times around his/her body. So the hammer goes round in a circle, with the person swinging it at the centre. When he/she let’s go of the cable, the hammer flies along a tangent of the circle – in a straight line. The object of the event is to throw the hammer as far as possible.

When the hammer is swinging in a circle, the man is pulling on the cable – he is exerting a force called the centripetal force. When he ceases to apply the centripetal force, the hammer no longer goes round in a circle – instead, it flies off in a straight line. So the centripetal force is required to maintain the hammer in circular motion.

In this rest of this blog, we shall think about the centripetal force in more detail. In the picture above, an object, like the hammer, is going round in a circle, centre at O, with a constant angular speed (see post 17.12) ω. The object starts at point P and at time t later it is at P’. From the definition of angular speed, the angle POP’ is ωt. We shall define the position of P’, relative to O, by the vector (see post 17.2) r.

If we use the line OP to define an X-axis, a Y-axis is generated at right angles to OP in the direction shown (see post 17.2). From the definition of sine and cosine (see post 16.50), the components of r in these directions are given by r.cos(ωt) and r.sin(ωt). (The dot “.” between r and sin or cos simply means “multiplied by” – just like the × sign.)

So we can write

r = ir.cos(ωt) + jr.sin(ωt)

as explained in post 17.2, where i and j are unit vector in the X and Y directions.

We can now calculate the velocity and acceleration of P and, hence, the centripetal force. But is order to do this, I need to tell you two results that I will justify in the Appendix to this post (parts 1.1 -1.3). One: if A and ω are constants, differentiation of Acos(ωt) with respect to time (see post 17.4) gives the result – ωAsin(ωt). Two: if A and ω are constants, differentiation of Asin(ωt) with respect to time (see post 17.4) gives the result ωAcos(ωt).

So, the velocity of P (see post 17.4) is given by

v = dr/dt = – iωr.sin(ωt) + jωr.cos(ωt).

Note that v is the translational velocity of P, defined in post 17.4, and not the angular velocity, defined in post 17.12. This expression for v also shows that its direction always makes an angle of 90o (π/2 rad) with r (see Appendix part 2). As a result, if the centripetal force is removed, P will move off in a tangent to the circle; the tangent is a line perpendicular to the radius that touches the circle at a single point.

Similarly, the acceleration of P (see post 17.4) is given by

a = dv/dt = – iω2r.cos(ωt) – jω2r.sin(ωt) = – ω2[ir.cos(ωt) + jr.sin(ωt)].

Note that a is the translational acceleration of P, defined in post 17.4, and not the angular acceleration, α, defined in post 17.12.

Comparison of the expressions for r and a shows that

a = – ω2r.

This means that an object at P’ that is going round a circle whose centre is at O, must be undergoing a translational acceleration along P’O – that is, it is accelerating towards the centre of the circle.

Now let’s suppose that the object has a constant mass m. According to post 16.13, it must be experiencing a force of

F = ma = – mω2r.

The force F represents the centripetal force required to keep the object moving in a circle. It is directed in the opposite direction to r (as shown by the minus sign in the equation) and so acts towards the centre of the circle. The concept of centripetal force is essential for understanding why objects, like satellites, move in a circular path. It is very important not to confuse centripetal and centrifugal forces; centrifugal forces are forces observed in a rotating frame of reference (see post 16.9) and will be the subject of a future post.

Related posts

Appendices: some mathematical results

1.1. Differentiation of Asin(t) with respect to t gives Acos(t)

I’m not going to prove this result (see part 1 of the Appendix of post 16.50 to find out more about mathematical proof); I’ll just show that it looks reasonable.

Remember, that when we differentiate something, we find the slope of a line at a point on a curve (post 17.4). In the graph above, Asin(t) is plotted (in red) against t. Now let’s consider the slope of this line at certain values of t and compare this slope with the value of Acos(t) plotted (in blue) on the same graph.

 t Slope of Asin(t) Value of Acos(t) 0 Maximum Maximum 90o 0 0 180o Maximum negative (downhill) Maximum negative (downhill) 270o 0 0 360o Maximum Maximum

We can see that the slope of Asin(t) is equal to the value of Acos(t).

So differentiation of Asin(t) yields Acos(t).

1.2 Differentiation of Asin(ωt) with respect to t gives ωAcos(ωt)

We will define φ (the Greek letter “phi”) as

φ = Asin(ωt),

where A and ω are constants, and u as

u = ωt.

So we can write φ as

φ = Asin(u)

Differentiation of φ with respect to u gives

dφ/du = Acos(u)

according to the previous section of this Appendix.

Let’s think about tiny changes in φ, u and t denoted by δφ, δu and δt (see post 17.4).

It seems reasonable to suppose that

(δφ/δt) = (δφ/δu) × (δu/δt).

In the limit as δt tends to zero (see post 17.4), this supposition becomes

(dφ/dt) = (dφ/du) × (du/dt).

We have already calculated /du. Using the results of part 2 of the Appendix to post 17.4, gives

du/dt = ω.

So

/dt = ωAcos(ωt).

1.3 Differentiation of Acos(ωt) with respect to t gives – ωAsin(ωt)

1. The angle between v and r

There are two ways in which we can multiply two vectors v and r; the first, called the dot product and written v.r, generates a scalar result and the second, called the cross product and written v × r, generates a vector result. (Note that there is only one way of multiplying something by a scalar, so then, for example, a.b means the same as a × b and a.b means the same as a × b; a scalar is a number with a value or magnitude but has no associated direction – post 16.50.)

The dot product of v and r is defined by

v.r = vr.cos θ

where θ is the angle between v and r. Here v and r have their usual meanings of the moduli of v and r, respectively.

Above we saw that

v = – iωr.sin(ωt) + jωr.cos(ωt).

Then the modulus (see post 17.2) of v is given by

v = √[ω2r2.sin2(ωt) + ω2r.cos2(ωt)] = ωr√ [sin2(ωt) + cos2(ωt)] = ωr.

The final result follows because the square of the sine of any angle added to the square of its cosine is 1 (post 17.2) and the square root of 1 is 1 (for a reminder about square roots see post 17.2). The modulus of r is simply r.

So

v.r = ωr2cos θ                                                          (1)

Remember that the cosine of zero is 1 and that the cosine of 90o (π/4 rad) is zero (see post 16.50). Now let’s think about the three vectors i, j, and k which have the magnitude equal to the number one and are mutually perpendicular (the angle between any two of them is 90o). It then follows, from the definition of v.r, that

i.j = i.k = j.i = j.k = k.i = k.j =0.

The angle between each of the vectors i, j, and k and itself must be zero, so that

i.i = j.j = k.k = 1.

So we can also write v.r in the form

v.r = [- iωr.sin(ωt) + jωr.cos(ωt)].[ir.cos(ωt) + jr.sin(ωt)]

= i.i[ωr2sin(ωt).cos(ωt)] + i.j[- ωr2sin2(ωt)] + j.i[ωr2cos2(ωt)]

+ j.j[ωr2cos(ωt).sin(ωt)]

= – [ωr2sin(ωt).cos(ωt)] + 0 + 0 + [ωr2sin(ωt).cos(ωt)] = 0.                                (2)

Now let’s compare equations 1 and 2. Since ω and r are not zero, the only way v.r can be zero (when we think of an object going round in a circle, as shown in the second picture of this post) is if cos θ = 0; in other words, θ must be 90o (see post 16.50).