# Skill Set 05.1

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### 1. What is the period of a merry-go-round with a radius of 1.75 m if the centripetal acceleration at this point is 7.2 m/s/s? (3.1 s)

The formula for this problem is: $a = \frac{4 \pi\ ^2 r}{T^2}$

After you plug the numbers in, it's pretty straight forward to solve for the time (T) $7.2\ m/s/s = \frac{4 \pi\ ^2 (1.75\ m)}{T^2}$ $T^2 = \frac{4 \pi\ ^2 (1.75m)}{(7.2\ m/s/s)}$

Take the square root of both sides.. $T^2 = 9.59545$ $T = 3.1 s$

No problem yeh? 2. What is the velocity of a 1350 kg car going around a 213 m radius corner if the frictional force is 9830 N? (39.4 m/s) $f = \frac{(m)(v)^2}{r}$

Jus plug in your numbers into tha formula $9830N = \frac{(1350kg)(v)^2}{213m}$

Multiply by 213 $2093790Nm\ = (1350kg)(v)^2$

Divide by 1350 $1550.96\ = v^2$

Square root it baby $v\ = 39.4 m/s$

3. Tycho twirls a 2.7 kg hammer at 6.2 m/s around a 5.6 m radius circle. What force in what direction must he exert on the hammer at the top and at the bottom? (8.0 N up, and 45 N up)

Forrmmmuullaaas: $a = \frac {v^2}{r}$ $\ = ma$

Find tha acceleration please $a = \frac {6.2^2}{5.6} \ = 6.86$

Find tha force please $(2.7)(9.81)\ = 26.487$

Plug er in there $\ = (2.7)(-6.86)\ = 8N up$

^(accel = negative cuz it's at the top)^ $\ = (2.7)(6.86)\ = 45N up$

Oke 4. At what distance from the center of a 4.59 x 1024 kg planet is the gravitational force on a 15.0 kg object equal to 60.0 N? (8.75x106 m)

Use: $F = \frac{(G)(m_1)(m_2)}{r^2}$
Which equals.. $r^2 = \frac{(G)(m_1)(m_2)}{f}$

Plug your numerical information in $r^2 = \frac{(6.67 x 10^{-11})(15kg)(4.59 x 10^{24})}{60N}$ $r^2\ = 7.65 x 10^{13}$

Do yo thing $r\ = 8.75 x 10^6$

5. Your 12,500 kg spaceship is orbiting 1.16x107 m from the center of a planet every 17,500 s. What is the mass of the planet? (3.02x1024 kg)

Formula is $G \frac{m_1 \ m_2}{r^2} = \frac {m_14\pi^2 r}{t^2}$

Plug your junk in $6.67 * 10^{-11} \frac{(12500kg) \ m_2}{(1.16 * 10^7m)^2} = \frac {(12500kg)4\pi^2 1.16 * 10^7}{17500s^2}$ $6.67 * 10^{-11}{(12500kg) \ m_2} = ((1.16 * 10^7m)^2)(18691.822)$ $m_2\ = \frac{((1.16 * 10^7m)^2)(18691.822)}{6.67 * 10^{-11}(12500kg)}$ $m_2\ = 3.02 * 10^{24} kg$