Answer to Question #219019 in Classical Mechanics for Laura

Question

1. What centripetal force is required to cause a 12.0 kg object to complete seven revolutions around a horizontal circle of radius 7.50 m in 24.0 s?

2.A satellite is located 2.75 x 10^5 m above the surface of the Earth. What is the period of the satellite’s orbit?

3. The Earth is 1.49 x 10^11 m from the Sun. If the Earth requires 365.25 days to go once around the Sun, what is the centripetal force on the Earth?

4.A 47.0 kg student starts from rest and slides 26.0 m down a hill where the average force of friction is 55.0 N. What is the final speed of the student at the bottom of the hill if the vertical drop was 17.0 m?

5. A 4.0 kg ball is swung at a constant speed in a vertical circle of radius 2.5 m. If the ball completes one revolution in 3.0 s, what is the tension in the rope at the top of the circle?

Answer 1

$1.$

$m = 12kg;r=7.5m;n=7;t=24s$

$F = ma = m\frac{v^2}{r}$

$v =\frac{s}{t}=\frac{2\pi rn}{t}=\frac{2\pi *7.5*7}{24}=13.74$

$F = m\frac{v^2}{r}= 12*\frac{13.74^2}{7.5}=302.26N$

$\text{Answer:} 302.26N$

$2.$

$F = m_sa$

$F = G\frac{m_sm_e}{(R+h)^2};$

$\text{where }$

$G = 6.67*10^{-11} \text{- Gravitational constant}$

$R = 6.4*10^6 m-\text {radius of the earth}$

$m_e= 6*10^{24}kg\text{ mass of earth}$

$m_s \text{ mass of satellite }$

$m_sa = G\frac{m_sm_e}{(R+h)^2};$

$a = G\frac{m_e}{(R+h)^2};$

$a = 6.67*10^{-11}*\frac{6*10^{24}}{(6.4*10^6+2.75*10^5)^2}=8.98$

$a = \frac{v^2}{r+h}$

$v= \sqrt{a(r+h)}= \sqrt{8.98*(6.4*10^6+2.75*10^5)}=7743\frac{m}{s}$

$T = \frac{s}{v}= \frac{2\pi (r+h)}{v}= 5416.49s$

$\text{Answer: }T = 5416.49s$

$3.$

$m_e= 6*10^{24}kg\text{ mass of earth}$

$F = m_ea$

$a = \frac{v^2}{r}$

$v = \frac{s}{t}$

$t = 365.25 \text{ days}= 365.25*24*3600= 31557600 s$

$v = \frac{s}{t} = \frac{2\pi r}{t} = \frac{2\pi *1.49*10^{11}}{31557600 } = 29666.21\frac{m}{s}$

$a = \frac{v^2}{r}= \frac{29666.21^2}{1.49*10^{11}}=0.0059\frac{m}{s^2}$

$F = m_ea= 6*10^24*0.0059= 3.54*10^{22}N$

$\text{Answer: }3.54*10^{22}N$

$4.$

$\text{at the top of the hill}$

$E = mgh = 47*9.8*17=7830.2J$

$\text{at the foot of the hill}$

$E =F_fr*s+\frac{mv^2}{2}=26*55+23.5v^2=1430 +23.5v^2$

$1430 +23.5v^2 = 7830.2$

$v\approx 16.5 \frac{m}{s}$

$\text{Answer: }16.5\frac{m}{s}$

$5.$

$v = \frac{s}{t}= \frac{2\pi r}{t}=\frac{2\pi*2.5}{3}=5.23\frac{m}{s}$

$F = ma = m\frac{v^2}{r}=4*\frac{5.23^2}{2.5}=43.76N$

$F = mg+T$

$T = F -mg= 43.76-4*9.8=4.56N$

$\text{Answer:4.56N}$