please help with the circular pendulumQ (2 Viewers)

CcYann · Aug 19, 2012

tks in advance

CcYann · Aug 19, 2012

anyone?
please?

nightweaver066 · Aug 19, 2012

a) Resolving the forces of the mass:
Vertically: $T_1 cos\theta - T_2 cos\theta = mg = 20 ... (1)$

Horizontally: $T_1 sin\theta + T_2 sin\theta = \frac{mv^2}{r} = 72 ... (2)$

For $T_1, (1) \times sin\theta + (2) \times cos\theta ,$

$2T_1 sin\theta cos\theta = 20sin\theta + 72cos\theta$

You can evaluate sin\theta and cos\theta as you know the radius is 0.5m and the hypotenuse of the two right angle triangles are 2m

Repeat similar steps for $T_2$ ,

$2T_2 sin\theta cos\theta = 72cos\theta - 20sin\theta$

b) Let's resolve the horizontal component of force but keep the velocity a variable,

$T_1 sin\theta + T_2 sin\theta = \frac{2v^2}{0.5} = 4v^2 ... (3)$

Find $T_2$ using this, so $(3) \times cos\theta - (1) \times sin\theta,$

$2T_2 sin\theta cos\theta = 4v^2 cos\theta - 20 sin\theta$

Evaluate $sin\theta$ and $cos\theta$ , similar to part (a), and for T_2 to exist, T_2 > 0. Then solve for v.

asianese · Aug 19, 2012

Looks like TSFX to me

CcYann · Aug 19, 2012

asianese said:
Looks like TSFX to me

yes, it is.
but the answer is a bit confuse me

CcYann · Aug 19, 2012

thank u so much =)

nightweaver066 said:
a) Resolving the forces of the mass:
Vertically: $T_1 cos\theta - T_2 cos\theta = mg = 20 ... (1)$

Horizontally: $T_1 sin\theta + T_2 sin\theta = \frac{mv^2}{r} = 72 ... (2)$

For $T_1, (1) \times sin\theta + (2) \times cos\theta ,$

$2T_1 sin\theta cos\theta = 20sin\theta + 72cos\theta$

You can evaluate sin\theta and cos\theta as you know the radius is 0.5m and the hypotenuse of the two right angle triangles are 2m

Repeat similar steps for $T_2$ ,

$2T_2 sin\theta cos\theta = 72cos\theta - 20sin\theta$

b) Let's resolve the horizontal component of force but keep the velocity a variable,

$T_1 sin\theta + T_2 sin\theta = \frac{2v^2}{0.5} = 4v^2 ... (3)$

Find $T_2$ using this, so $(3) \times cos\theta - (1) \times sin\theta,$

$2T_2 sin\theta cos\theta = 4v^2 cos\theta - 20 sin\theta$

Evaluate $sin\theta$ and $cos\theta$ , similar to part (a), and for T_2 to exist, T_2 > 0. Then solve for v.

but if it's possible to say theta is 30 degree because the radius is 0.5?

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please help with the circular pendulumQ (2 Viewers)

CcYann

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CcYann

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nightweaver066

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asianese

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CcYann

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CcYann

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