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teratoma teratoma
wrote...
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12 years ago
A child ties a rock to a string and whirls it around in a horizontal circle. Assuming a 1.18m -long string making 30.0 (degree) angle below the horizontal, find the speed of the rock.

PART A
Express your answer with the appropriate units.

v = ?

my answer 1.4 m/s (wrong)

PART B


Find the period of its uniform circular motion.

Express your answer with the appropriate units.

T = value?, units?
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wrote...
Valued Member
12 years ago
Rotation radius (perpendicular to the turning axis):

r = 1.18 x cos(30) .. .. r = whatever you get... let's call it r since I don't have a calculator Face with Rolling Eyes

Tension in string:

Vertical component = mg ... (supporting the weight of the rock)
Horizontal component = mv2/ r ... (providing the centripetal force)

So:

tan(30) = mg \(/\) (m * v2 / r) ...

tan(30) = g*r / v2

v2 = g*r / tan(30) ... .. 9.80 x 1.18 / tan(30) .. .. v2 = some number (m/s)2 ...

Period time = circumference / velocity .. .. (2\({\pi}\) x 1.18) / 4.70 = your answer.
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wrote...
12 years ago
Consider this question:

A child ties a rock to a string and whirls it around in a horizontal circle. Assuming a 1.38m-long string making 20.0degrees angle below the horizontal, find the speed of the rock.

I'm assuming it's very similar to yours!

Here's the solution

Consider the tention force of the atring as T, and mass of the object is m.

It is very helpfull to you if I can draw a figure.. but it is difficult to draw a figure here.

Consider the the object and resolve all froces in radial direction, then, Centrafigual force of mgV^2/r is acting to outer direction and T*Cos20 to inword direction

Rightwards Arrow TCos20 = mV^2/r                     ...1

Consider the vertical forces

Rightwards Arrow TSin20 = mg                     ...2

from 2/1 Rightwards Arrow Tan20 = g*r/V^2

r = 1.38 Cos20

V = 5.91 m/s

Enjoy
could of, should of, would of...
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