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Gnoma [55]
3 years ago
7

A green hoop with mass mh = 2.4 kg and radius Rh = 0.14 m hangs from a string that goes over a blue solid disk pulley with mass

md = 2.3 kg and radius Rd = 0.08 m. The other end of the string is attached to an orange block on a flat horizontal surface that slides without friction and has mass m = 3.6 kg. The system is released from rest.
(a) What is magnitude of the linear acceleration of the hoop?
(b) What is magnitude of the linear acceleration of the block?
(c) What is the magnitude of the angular acceleration of the disk pulley?
(d) What are the tensions in the string between the block and disk pulley and between the hoop and disk pulley?
(e) What is the speed of the green hoop after it falls distance d = 1.49 m from rest.
(f) Now use work-energy conservation to solve for the hoop’s speed in (e).
(g) Now instead of the block, the other end of the string is attached to a massless axel through the center of an orange sphere that rolls without slipping and has mass m = 3.6 kg and radius R = 0.22 m (Figure 2). Use energy conservation principle to solve for the speed of the hoop after it falls distance d = 1.49 m from rest.

I'm completely lost. I know that the weight of the green hoop is the only force acting on our system. Therefore: F_{net} = m_{h}g = 23.544 Newtons
For part one, I know \alpha_{n}=\frac{a}{R_{n}}. However, I'm not sure how to find \alpha_{n}.

Any help is appreciated.

Physics
1 answer:
sineoko [7]3 years ago
8 0

Answer:

a). linear acceleration of the hoop and block =  3.2895 <u>m</u>

                                                                                            s²

c). magnitude of the angular acceleration of the disk pulley = 41.119 <u>rad</u>

                                                                                                                  s²

d). tensions in the string between the block and disk pulley = 11.842 N

     tensions in the string between  the hoop and disk pulley = 15.625 N

check the pictures below for further explanation and for the remaining answers. I hope it helps you. Thank you

Explanation:

Start by writing "F=ma" equations for each of the things that moves. Also, since some of the objects (the pulley and the orange sphere) rotate, you should write "τ = Iα" equations (net torque = moment of inertia × angular acceleration) for those. In the end, you should have enough equations that you can combine them and solve for the desired quantities.

First, the hoop. There's no indication that it rotates, so we don't need a "τ = Iα" equation for it; just do "F=ma". The hoop has gravity ((mhoop)g) pulling down, and the tension in the vertical string (call it "T_v") pulling up.

Fnet = ma

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Explanation:

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tan^{-1}(\frac{11.1}{-43.0})=-14.5 but since we are in Q2 we have to add 180 degrees to that angle giving us 165.5 degrees

8 0
3 years ago
Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley
kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = \frac{g}{7}

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = \frac{4m}{2} x^2  + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2

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also the general potential energy can be expressed as

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The Lagrangian of the problem can now be setup as

L =4mx^2 +2my^2 -2mxy +2mgy + constant

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange  equation = 4x-y =0\\-2y+x +g = 0

solving the equations simultaneously

x ( acceleration of mass 4m ) = \frac{g}{7}

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Answer:

For left = 0  N/C

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Explanation:

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Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

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A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

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Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

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\Delta p_x= 6.6251 \,kg.m.s^{-1}

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\Delta p_y= 7.65\times sin 30^{\circ}

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also, impulse

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where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

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F_x=144.3396\times cos 30^{\circ}

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&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
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