Question

Pre-Lab 7: Centripetal force using a pendulum Name: 1. ill of the Jungle swings on a vine 4.00m+0.01m long. What is the tension in the vine at in the the·bottom ,of her swing if Jill, whose mass is 500kg,100kg is. moving at 3.0m st0.05ms's when the vine is vertical.(The vine is vertical at the lowest point swingt) A free body diagram will be useful here! Show work. Free Body Force Diagram: Newton's 23d law equation (symbols only): Solve for Tension and plug in numbers FTension

Answer 1

1.

m = 50.0 plusminus 1.00 kg v = 3.00 plusminus 0.05 m/s L = 4.00 plusminus 0.01 m F_T - mg = mv^2/L F_T = mg + mv^2/L F_T = 50.0 * 9.81 + 50.0 * (3.00)^2/(4.00) = 603 N Let A = mg and B = mv^2/L sigma^2_F_T = sigma^2_A + sigma^2_B sigma^2_F_T = g^2 sigma^2_m + (mv^2/L)[(sigma_m/m)?^2 + (2 sigma_v/v)^2 + (sigma_L/L)^2] sigma^2_F_T = (9.81)^2(1.00)^2 + (50.0 * (3.00)^2/(4.00))^2[(1.00/50.0)^2 + (2 * 0.05/3.00)^2 + (2 * 0.05/3.00)^2 + (0.01/4.00)^2] sigma_F_T = 11 F_T = 603 N plusminus 11N 2 Length of string L = 2.000m Radius of circle R = L sin theta = 100.0 cm Time period t = 2.600 s Since net vertical force is zero, T cos theta = mg ---(1) Net horizontal force provides the centripetal acceleration, T sin theta = mv^2/R ---(2) Time period t = 2.600s = 2 pi R/v Hence speed of mass is v = 2 pi R/t = 2 pi * 1.000/2.600

Let and

$\sigma_{F_T}^2=\sigma_A^2+\sigma_B^2$

$\sigma_{F_T}^2=g^2\sigma_m^2+\left ( \frac{mv^2}{L} \right )^2\left [ \left ( \frac{\sigma_m}{m} \right )^2+\left ( \frac{2\sigma_v}{v} \right )^2+\left ( \frac{\sigma _L}{L} \right )^2 \right ]$

2.

Length of string $L=2.000\,m$

Radius of circle  

Time period $t=2.600\,s$

Since net vertical force is zero, $T\cos\theta=mg$ ------ (1)

Net horizontal force provides the centripetal acceleration, ------(2)

Time period $t=2.600\,s=\frac{2\pi R}{v}$

Hence speed of mass is $v=\frac{2\pi R}{t}=\frac{2\pi *1.000}{2.600}$

From length and radius ,

Angle

Using equations (1) and (2)

$\tan\theta=\frac{v^2}{Rg}$

acceleration due to gravity

$g=\frac{v^2}{R\tan\theta}=\frac{\left(\frac{2\pi R}{t} \right )^2}{R\tan30\degree}=\frac{4\pi^2 R}{t^2\tan30\degree}=10.12\,m/s^2$