Skill Set 05.1

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Main Page > IB Physics Skill Sets > Skill Set 05.1

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? Eman.jpg

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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

Not an issue Eman.jpg

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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}

 <F_1-F_2>\ = 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

 <F_1-26.487>\ = (2.7)(-6.86)\ = 8N up

^(accel = negative cuz it's at the top)^

 <F_1-26.487>\ = (2.7)(6.86)\ = 45N up

Oke Eman.jpg

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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

Yeh Eman.jpg

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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

Git-R-Done Eman.jpg

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