Question

shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.040 kg, the string has length L = 0.90 m and negligible mass, and the bob follows a circular path of circumference 0.94 m. What are (a) the tension in the string and (b) the period of the motion?

Answer 1

a)
`T=0.40 N`

b)
`T=1.9 s`

Step-by-step explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

`P=2\pi r=0.94`

`r=(0.94)/(2\pi)=0.15 m`

Now, we need to find the angle of the pendulum from vertical.

`tan(\alpha)=(r)/(L)=(0.15)/(0.90)=0.17`

`\alpha=9.44 ^(\circ)`

Let's apply Newton's second law to find the tension.

`\sum F=ma_(c)=m\omega^(2)r`

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

`Tcos(\alpha)=mg` (1)

The horizontal equation of motion will be:

`Tsin(\alpha)=m\omega^(2)r` (2)

a) We can find T usinf the equation (1):

`T=\frac {mg}{cos(\alpha)}=(0.04*9.81)/(cos(9.44))=0.40 N`

We can find the angular velocity (ω) from the equation (2):

`\omega=\sqrt{(Tsin(\alpha))/(mr)}=3.31 rad/s`

b) We know that the period is T=2π/ω, therefore:

`T=(2\pi)/(\omega)=(2\pi)/(3.31)=1.9 s`

I hope it helps you!