The formula for centripetal force is:

\(\displaystyle F_c=ma_c=\frac{mv^2}{r}\)

While we can find what the acceleration will be, there is no way for us to find the mass. The force cannot be calculated.

The formula for centripetal acceleration is:

\(\displaystyle a_c=\frac{v^2}{r}\)

We are given values for the linear velocity and radius, allowing us to solve for the centripetal acceleration.

\(\displaystyle a_c=\frac{(0.02\frac{m}{s})^2}{2.1m}\)

\(\displaystyle a_c=\frac{0.0004\frac{m^2}{s^2}}{2.1m}\)

\(\displaystyle a_c=0.00019\frac{m}{s^2}\)

The formula for centripetal force is:

\(\displaystyle F_c=\frac{mv^2}{r}\)

We are given the girl's mass, her linear velocity, and the radius of her motion. Using these values, we can calculate the centripetal force.

\(\displaystyle F_c=\frac{(41kg)(0.02\frac{m}{s})^2}{2.1m}\)

\(\displaystyle F_c=\frac{(41kg)(0.0004\frac{m^2}{s^2})}{2.1m}\)

\(\displaystyle F_c=0.0078N\)

The equation for torque is:

\(\displaystyle \tau=F*r\)

We are given the force and distance (radius), allowing us to solve for the torque by multiplication.

\(\displaystyle \tau=13.2N*1.1m\)

\(\displaystyle \tau=14.52N*m\)

The formula for torque is:

\(\displaystyle \tau=F*r\)

We are given the force applied and the torque generated, allowing us to solve for the length of the lever arm.

\(\displaystyle 16N*m=(18N)r\)

\(\displaystyle \frac{16N*m}{18N}=r\)

\(\displaystyle 0.89m=r\)

The formula for torque is:

\(\displaystyle \tau=F*r\)

We are given the length of the lever arm and the torque generated, allowing us to solve for the force applied.

\(\displaystyle 11N*m=F(0.43m)\)

\(\displaystyle \frac{11N*m}{0.43m}=F\)

\(\displaystyle 25.58N\)

To solve this problem, recognize that the force due to friction must equal the centripetal force of the curve:

\(\displaystyle F_f=F_c\)

This will give the maximum force that the car can have in the curve without skidding. Expand the equation for centripetal force.

\(\displaystyle F_f=m\frac{v^2}{r}\)

We are given the value for the force of friction, the mass of the car, and the radius of the curve. Using these values, we can find the velocity.

\(\displaystyle 2587.2N=1200kg*\frac{v^2}{4.2m}\)

\(\displaystyle \frac{2587.2N}{1200kg}=\frac{v^2}{4.2m}\)

\(\displaystyle 2.156\frac{m}{s^2}=\frac{v^2}{4.2m}\)

\(\displaystyle 4.2m*2.156\frac{m}{s^2}=v^2\)

\(\displaystyle 9.0552\frac{m^2}{s^2}=v^2\)

\(\displaystyle \sqrt{9.0552\frac{m^2}{s^2}}=\sqrt{v^2}\)

\(\displaystyle 3.01\frac{m}{s}=v\)

The equation for centripetal force is:

\(\displaystyle F_c=m\frac{v^2}{r}\)

We are given the radius (length of the rope), velocity, and mass of the rock, allowing us to calculate the centripetal force.

\(\displaystyle F_c=0.22kg*\frac{(0.9\frac{m}{s})^2}{0.8m}\)

\(\displaystyle F_c=0.22kg*\frac{0.81\frac{m^2}{s^2}}{0.8m}\)

\(\displaystyle F_c=0.22kg*1.0125\frac{m}{s^2}\)

\(\displaystyle F_c=0.223N\)

Angular velocity is given by the equation:

\(\displaystyle \omega=\frac{v}{r}\)

We know the radius, but we need to find the tangential velocity. We can do this be finding the centripetal acceleration from the centripetal force.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force acting on the satellite. That means \(\displaystyle F_c=F_G\).

Solve for \(\displaystyle F_G\) for the satellite. To do this, use the law of universal gravitation.

\(\displaystyle F_G=G\frac{m_1m_2}{r^2}\)

Remember that \(\displaystyle r\) is the distance between the centers of the two objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the masses of the objects and distance to solve for the force of gravity.

\(\displaystyle F_G=G\frac{m_sm_E}{(r_E+h)^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}\)

\(\displaystyle F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})\)

\(\displaystyle F_G=6.87N\)

Now that we know the force, we can find the acceleration. Remember that centripetal force is \(\displaystyle F_c=m*a_c\). Set our two forces equal and solve for the centripetal acceleration.

\(\displaystyle F_G=F_c=m_s*a_c\)

\(\displaystyle 6.87N=m_s*a_c\)

\(\displaystyle 6.87N=2500kg*a_c\)

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}=a_c\)

Now we can find the tangential velocity, using the equation for centripetal acceleration. Again, remember that the radius is equal to the sum of the radius of the Earth and the height of the satellite!

\(\displaystyle a_c=\frac{v^2}{r}\)

\(\displaystyle 2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}\)

\(\displaystyle (2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}\)

\(\displaystyle 2.85*10^{15}\frac{m^2}{s^2}=v^2\)

\(\displaystyle \sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}\)

\(\displaystyle 5.33*10^7\frac{m}{s}=v\)

Now that we know the tangential velocity, we can divide by the radius to find the angular velocity. Again, remember that the radius of the orbit is equal to the sum of the Earth's radius and the height of the satellite above the surface.

\(\displaystyle \omega=\frac{v}{r}\)

\(\displaystyle \omega=\frac{5.33*10^7\frac{m}{s}}{1.45*10^{17}m}\)

\(\displaystyle \omega=3.68*10^{-10}\frac{rad}{s}\)

The given equation for moment of inertia is:

\(\displaystyle I_{sphere}=\frac{2}{5}mr^2\)

Use the given values for the mass and radius of the ball to solve for the moment of inertia.

\(\displaystyle I=\frac{2}{5}mr^2\)

\(\displaystyle I=\frac{2}{5}(1.5kg)(0.038m)^2\)

\(\displaystyle I=8.7*10^{-4}kg*m^2\)