Question

2. [2] A mass m 8.100 ky is suspended from a string of length vertical (in degrees)? 1.150 m. It revolves in a horirontal circle (see Figure). The tangential speed of the mass is 2.403 ... What is the angle theta between the string and the 3.12pt] Calculate, using Newton's law of gravity, the magnitude of the force of attraction between the Earth and a mass of 33.000 kg on the Earth's surface. The distance to the center of the Earth from the surface is 6370 km. The mass of the Earth is 5.98 x 10kg. Gravitational constant G 667 x 10 Nm /kg?. S A A Last Answer: 9.03 N Hint: The SI system of units uses metres as the unit of length. You must change the distance from k tom. How does your answer compare with the value of me?

Answer 1

here,

mass , m = 8.1 kg

length , l = 1.15 m

radius of the circle , r = l * sin(theta)

r = 1.15 * sin(theta)

tangential speed , v = 2.403 m/s

let the tension in the string be T and angle be theta

equating the forces vertically

T * cos(theta) = m * g .....(1)

and

equating the forces horizontally

T * sin(theta) = m * v^2 /r ....(2)

from (1) and (2)

tan(theta) = v^2 /rg

tan(theta) = 2.403^2 /(1.15 * sin(theta) * 9.81)

solving for theta

theta = 38.9 degree

the angle with vertical is 38.9 degree

3)

radius , r = 6370 km = 6.37 * 10^6 m

mass of Earth , m1 = 5.98 * 10^24 kg

m2 = 33000 kg

the magnitude of force , F = G * m1 * m2 /r^2

F = 6.67 * 10^-11 * 33000 * 6.37 * 10^6 /(6.37 * 10^6)^2 N

F = 0.35 N