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mezya [45]
3 years ago
9

A 5 kg uniform disk with a 40 cm radius swings without friction about a nail through a point 10 cm from the rim. It is released

from rest from a position with its center above the nail.(a) What is its angular velocity when the center is directly below the nail?(b) What is its angular acceleration when the center is level with the nail?(c) Suppose it hangs with its center directly below the nail. If it is pushed a small distance from this equilibrium position then what is its period of small oscillations?
Physics
1 answer:
xz_007 [3.2K]3 years ago
8 0

(a) Angular velocity is 8.31 rad/s

(b) Angular acceleration is 17.29 rad/s²

(c) Period of oscillation is 1.51 sec

<u>Explanation</u>

Given:

Mass, m = 5 kg

Radius, r = 40 cm

Distance, x = 10 cm

(a)

Angular velocity, ω = ?

In this case, energy is conserved.

\frac{1}{2} I w^2 = mg (2a)\\\\w^2 = \frac{4mag}{I} \\\\w = \sqrt{\frac{4mag}{I} }

where,

I = moment of Inertia

g = gravity

a = distance between the nail and the center

To find I of nail:

I_n_a_i_l = I_c_e_n_t_e_r + ma^2

and

I_c_e_n_t_e_r = \frac{mr^2}{2} \\\\I_c_e_n_t_e_r = \frac{5 X 0.4^2}{2} \\\\I_c_e_n_t_e_r = 0.4 kg.m^2

On substituting the value:

I_n_a_i_l = 0.4 + 5 ( 0.4 - 0.1) ^2\\\\I_n_a_i_l = 0.4 + 0.45\\\\I_n_a_i_l = 0.85 kg.m^2

So, ω is:

w = \sqrt{\frac{4 X 5X 9.8 X 0.3}{0.85} } \\\\w = 8.31 rad/s

(b)

Angular acceleration, α = ?

We know:

\alpha = \frac{t}{I}

where,

ζ = torque

I = moment of Inertia

To calculate torque:

t = F X a\\  \\  = mg X a\\\\= 5 X 9.8 X 0.3\\\\= 14.7 Nm

Substituting the value we get:

\alpha  = \frac{14.7}{0.85} \\\\\alpha  = 17.29 rad/s^2

Thus, the angular acceleration is 17.29 rad/s²

(c)

Period of oscillation, T = ?

We know:

T = 2\pi \sqrt{\frac{I}{mga} }

On substituting the value:

T = 2 X 3.14 X\sqrt{\frac{0.85}{5 X 9.8X0.3} } \\\\T = 1.51 s

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A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def
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Answer:

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Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

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V2=0.283m/s

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Question:
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Answer:

It states a fact about how nature works.

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WILL GIVE BRAINIEST TO CORRECT ANSWER
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Answer:

<em>Hey mate, here's ur answer</em>

<em>-------------------------------------------------------------</em>

<u><em>Loudness</em></u><em> refers to how a sound seems to a listener, whether it's loud or soft.  </em>

<em>___________________________</em>

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<em>Hope this helps</em>

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