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

Answer:

a) $T=0.40 N$

b) $T=1.9 s$

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:

$P=2\pi r=0.94$

$r=\frac{0.94}{2\pi}=0.15 m$

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

$tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17$

$\alpha=9.44 ^{\circ}$

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):

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

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

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

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

$T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s$

I hope it helps you!