2 Consider a conical pendulum (Fig.~P6.8) with a bob of mass
3 $m=80.0\U{kg}$ on a string of length $L=10.0\U{m}$ that makes an angle
4 of $\theta=5.00\dg$ with the vertical. Determine \Part{a} the
5 horizontal and vertical components of the force exerted by the string
6 on the pendulum and \Part{b} the radial acceleration of the bob.
13 draw((0,0,2u)--(0,0,0), dashed);
14 draw((0,0,2u)--(0,u,0));
15 draw(scale3(u)*unitcircle3);
24 Drawing a free body diagram for the bob
34 Angle t = Angle(dir(90), (0,0), dir(90+theta), "$\theta$");
36 Vector T = Force((0,0), mag=mg/Cos(theta), dir=(90+theta), "$T$");
38 Vector G = Force((0,0), mag=mg, dir=(-90), "$mg$");
42 draw_ijhat((.5u,0), idir=0);
46 The bob does not move up or down, so the sum of forces in the vertical
47 direction must be zero.
49 0 &= \sum F_y = T_y - mg \\
50 T\cos(\theta) = T_y &= mg
51 = 80.0\U{kg}\cdot9.80\U{m/s$^2$} = \ans{784\U{N}} \\
52 T &= \frac{mg}{\cos(\theta)} \;,
54 where I've marked the vertical component of the string tension. The
55 horizontal (radial) component is given by
58 = -\frac{mg\sin(\theta)}{\cos(\theta)}
59 = -mg\tan(\theta) = \ans{-68.6\U{N}} \;.
63 The radial acceleration is given by Newton's second law
66 a_r &= \frac{F_r}{m} = \frac{T_x}{m}
67 = \frac{-mg\tan(\theta)}{m}
69 = -9.80\U{m/s$^2$}\cdot\tan(5.00\dg)
70 = \ans{0.857\U{m/s$^2$}} \;.