Knowing the centripetal force on the mass and the radius of the circle, we can calculate its velocity:

$\displaystyle F_c = ma_c = m\frac{v^2}{r}$

Rearranging for velocity:

$\displaystyle v = \sqrt{\frac{F_cr}{m}} = \sqrt{\frac{(50N)(5m)}{10kg}}=5\frac{m}{s}$

We can use this to find the period of the mass:

$\displaystyle velocity = \frac{circumference}{period}$

Rearranging for period, we get:

$\displaystyle P = \frac{C}{v}=\frac{2\pi r}{v}=\frac{2\pi(5m)}{5\frac{m}{s}} = 2\pi s$

A normal clock registers 60 seconds for every rotation of the minute hand. We need to determine the new period per rotation.

A normal clock registers the following amount of seconds every day:

$\displaystyle 24hr*\frac{60min}{hr}*\frac{60s}{min}= 86400 s$

Losing 1500 seconds every day, we now have:

$\displaystyle 86400-1500=84900 s$

The hand in question completes one rotation every minute, so we'll dive this new time by minutes per day:

$\displaystyle (1day)(\frac{24hr}{day})(\frac{60min}{hr})= 1440 min$

Now dividing our two values to get seconds per rotation:

$\displaystyle P=\frac{84900}{1440}=59.0 s$

To start, we need to determine what exactly we are solving for. What does it mean for the string to be in tension at all points? This means that at some point in the circle, tension will equal zero; thus the force of gravity and the centripetal force will equal each other.

$\displaystyle F_c = F_g$

Expand these force expressions and simplify:

$\displaystyle ma_c = mg$

$\displaystyle a_c = g$

The expression for centripetal acceleration is:

$\displaystyle a_c = \frac{v^2}{r} = g$

We know $\displaystyle r$ (the length of the string) and $\displaystyle g$, but we need to develop an expression for velocity:

$\displaystyle velocity = \frac{circumference}{period} = (circumference)(frequency)$

$\displaystyle C = 2\pi r$

$\displaystyle v = Cf = (2\pi r)f$

Substituting this back into the equation for centripetal acceleration, we get:

$\displaystyle a_c = \frac{(2\pi r f)^2}{r} = 4\pi^2 rf^2 = g$

Rearrange for frequency:
$\displaystyle f = \sqrt{\frac{g}{4\pi^2 r}}$

We know all of these values, allowing us to solve:
$\displaystyle f = \sqrt{\frac{10\frac{m}{s^2}}{4\pi^2(0.5m)}} = 0.71 Hz$

Since the man is in space, the only force we have to worry about in this problem is the centripetal force, which results from the tension in the spring. Therefore, we are being asked what period gives us a centripetal force of $\displaystyle 125N$.

$\displaystyle F_c = ma_c = m\frac{v^2}{r}$

We need an expression for velocity:

$\displaystyle velocity = \frac{circumference}{period}$

$\displaystyle v = \frac{C}{P} = \frac{2\pi r}{P}$

Substitute this back into the original expression:

$\displaystyle F_c = m\frac{(\frac{2\pi r}{P})^2}{r} = \frac{4m\pi^2r}{P^2}$

Rearranging for period we get:

$\displaystyle P = \sqrt{\frac{4m\pi^2r}{F_c}}$

We know all of these values, allowing us to solve:
$\displaystyle P = \sqrt{\frac{4(5kg)(\pi^2)(1m)}{125N}} = 1.3s$

The ordinary frequency is the number of cycles per second. Since a second hand makes one revolution, or cycle, every 60s, the correct answer is $\displaystyle \frac{1}{60s}=0.01\bar{6}Hz$. You can also think of the ordinary frequency as the angular velocity divided by $\displaystyle 2\pi$.

We can begin with the conservation of energy to solve this problem:

$\displaystyle E = U_i + K_i = U_f +K_f$

The problem statement tells us that the cylinder is initially at rest, so we can eliminate initial kinetic energy. If we assume that the height of the cylinder when it reaches a period of 0.2s has a height of 0, we can eliminate final potential energy. Therefore, we get:

$\displaystyle U_i = K_f$

Expanding these terms and making sure we have both a linear and rotational component to kinetic energy, we get equation (1):

$\displaystyle mgh_i = \frac{1}{2}mv_f^2+\frac{1}{2}Iw_f^2$

Before we move on, we know that we are going to have to calculate something that we can use to determine the period of the cylinder. We know that the period is how long it takes the cylinder to complete one full rotation. Thinking practically, we can use the circumference of the cylinder and linear velocity to determine period:

$\displaystyle P = \frac{circumference}{velocity}$

Using variables, we get equation:

$\displaystyle P = \frac{2\pi r}{v_f}$

Rearranging for final velocity, we get equation (2):

$\displaystyle v_f = \frac{2\pi r}{P}$

Now we know that the period is dependent on final linear velocity. We will come back to this equation. Now we can go back to equation (1) and begin substituting in expressions for unknown variables moving from left to right. The first variable we don't know is initial height. However, we can use the distance the cylinder traveled and the angle of the slope:

$\displaystyle sin(a)=\frac{h_i}{l}$

Rearranging for initial height, we get equation (3):

$\displaystyle h_i = lsin(a)$

Moving on, the next unknown term is final velocity. We can substitute equation (2) that we already derived:

$\displaystyle v_f = \frac{2\pi r}{P}$

Moving on, the next unknown term is the moment of inertia. Using the expression for a cylinder to get equation (4):

$\displaystyle I = \frac{1}{2}mr^2$

Moving on, the final unknown term is final rotational velocity. We can use the relationship between this and linear velocity:

$\displaystyle w_f = \frac{v_f}{r}$

Now substituting equation (2), we get equation (5):

$\displaystyle w_f = \frac{\left ( \frac{2\pi r}{P}\right )}{r} = \frac{2\pi}{P}$

We can now substitute equations 2, 3, 4, and 5 into equation (1):

$\displaystyle mglsin(a) = \frac{1}{2}m\left ( \frac{2\pi r}{P}\right )^2+\frac{1}{2}\left ( \frac{1}{2}mr^2\right )\left ( \frac{2\pi}{P}\right )^2$

Eliminating mass from both sides of the equation and expanding each term:

$\displaystyle glsin(a) = \frac{2\pi ^2r^2}{P^2}+\frac{\pi ^2r^2}{P^2}$

Combining the terms on the right:

$\displaystyle glsin(a) = \frac{3\pi ^2r^2}{P^2}$

Rearranging for length:

$\displaystyle l = \frac{3\pi ^2r^2}{P^2gsin(a)}$

Check your units and make sure you end up with seconds before moving on!

We know values for each variable, so time to plug and chug:

$\displaystyle l = \frac{3\pi ^2(0.5m)^2}{(0.2s)^2\left ( 10\frac{m}{s^2}\right )sin(40^{\circ})}$

$\displaystyle l = 28.8m$

Finding distance of lap of inner car:

$\displaystyle 2\pi*450m=2826 m$

Convert $\displaystyle \frac{km}{hr}$ to $\displaystyle \frac{m}{s}$:

$\displaystyle 150\frac{km}{hr}*\frac{1000m}{1 km}*\frac{1hr}{3600s}=41.7\frac{m}{s}$

Use the distance formula:

$\displaystyle v=\frac{d}{t}$

$\displaystyle 41.7=\frac{2826}{t}$

$\displaystyle t=67.8s$

Inner car time:

Finding distance of lap of inner car:

$\displaystyle 2\pi*450m=2826 m$

Convert $\displaystyle \frac{km}{hr}$ to $\displaystyle \frac{m}{s}$:

$\displaystyle 150\frac{km}{hr}*\frac{1000m}{1 km}*\frac{1hr}{3600s}=41.7\frac{m}{s}$

Use the distance formula:

$\displaystyle v=\frac{d}{t}$

$\displaystyle 41.7=\frac{2826}{t}$

$\displaystyle t_1=67.8s$

Finding distance of lap of outer car:

$\displaystyle 2\pi*452m=2839 m$

Convert $\displaystyle \frac{km}{hr}$ to $\displaystyle \frac{m}{s}$:

$\displaystyle 150\frac{km}{hr}*\frac{1000m}{1 km}*\frac{1hr}{3600s}=41.7\frac{m}{s}$

Use the distance formula:

$\displaystyle v=\frac{d}{t}$

$\displaystyle 41.7=\frac{2839}{t}$

$\displaystyle t_2=68.1s$

$\displaystyle t_2-t_1=68.1-67.8=.3s$

$\displaystyle 50\frac{m}{s}*\frac{1 rotation}{2*\pi*.25m}*\frac{2\pi radians}{rotation}=200\frac{radians}{second}$

First, find the angular frequency:

$\displaystyle 50\frac{m}{s}*\frac{1 rotation}{2*\pi*.25m}*\frac{2\pi radians}{rotation}=200\frac{radians}{second}$

Angular period is the inverse of angular frequency:

$\displaystyle \frac{1}{200\frac{radians}{second}}=.005\frac{seconds}{radian}$