Question
1.Consider the motion of the rock in the figure below. What is the minimum speed the rock can have without the string becoming "slack"? (The rock is traveling in vertical circle. Assume that m = 1.3kg and r = 0.47 m.)
m/s
magnitude N direction Consider the motion of the rock in the figure below. What is the minimum speed the rock can have without the string becoming "slack"? (The rock is traveling in vertical circle. Assume that m = 1.3kg and r = 0.47 m.) Three lead balls of mass m1 = 14 kg, m2 = 26 kg, and m3 = 8.6 kg are arranged as shown in the figure below. Find the total gravitational force exerted by balls 1 and 2 on ball 3. Be sure to give the magnitude and the direction of this force. The region of the solar system between Mars and Jupiter contains many asteroids that orbit the Sun. Consider an asteroid in a circular orbit of radius 5.2 1011 m. Find the period of the orbit. Your weight is due to the gravitational attraction of the Earth. The Moon, though, also exerts a gravitational force on you, and when it is overhead, your weight decreases by a small amount. Calculate the effect of the Moon on your weight. Express your result as a percentage difference for the cases of the Moon overhead and the Moon on the opposite side of the Earth. A popular circus act features daredevil motorcycle riders encased in the "Globe of Death" (see the figure below), a spherical metal cage of diameter 17 ft. Assume a speed of 24 mi/h for both tricks. A rider of mass 63 kg on a 125-cc (95-kg) motorcycle keeps his bike horizontal as he rides around the "equator" of the globe. What coefficient of friction is needed between his tire and the cage to keep him in place? How many loops will the rider make per second? The same rider performs vertical loops in the globe. What force does the cage need to withstand at the top and the bottom of the rider's loop?
Explanation / Answer
1) m*v^2/r = m*g
v = sqrt(g*r) = 2.146 m/s
2)
r12 = sqrt(3^2+0.5^2) = 3.04 m
r23 = sqrt(1.5^2+1.5^2) = 2.12 m
F13 = G*m1*m3/r12^2 = 8.69*10^-10 N
theta = tan^-1(1/3) = 18.43 degrres
F13x = F13*cos(18.43) = 8.244*10^-10 N
F13y = -F13*sin(18.43) = -2.74*10^-10 N
F23 = G*m2*m3/r12^2 = 33.18*10^-10 N
theta = tan^-1(1.5/1.5) = 45 degrres
F23x = F13*cos(45) = 23.46*10^-10 N
F23y = F13*sin(45) = 23.46*10^-10 N
Fnetx = F13x + F23x = 31.7*10^-10 N
Fnety = F13y + F23y = 20.72*10^-10 N
Fnet = sqrt(Fnetx^2+Fnety^2) = 3.787*10^-9N
