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Ierofanga [76]
3 years ago
5

Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2

4 N, and F3 = 19 N. The radial distances are R2 = 0.22 m and R3 = 0.10 m. The forces F2 and F3 are perpendicular to the radial lines labeled R2 and R3. The moment of inertia of the cylinder is 12MR22. Find the magnitude of the angular acceleration of the cylinder about the axis of rotation.
Physics
1 answer:
Mila [183]3 years ago
6 0

Answer:

The angular acceleration is 11.66 rad/s²

Explanation:

Step 1: Given data

Three forces are applied to a solid cylinder of mass 12 kg

F1 = 15 N

F2 = 24 N

F3 = 19 N

R2 = 0.22m

R3 = 0.10m

Step 2: Find the magnitude of the angular acceleration

I = ½mr² = ½ * 12kg * (0.22m)² = 0.29 kg*m²

torque τ = I*α

τ = F2*R2 - F1*R1 = 24N*0.22m - 19N*0.10m = 3.38 N*m

This means

I = ½mr² = 0.29 kg*m²

τ = I*α  =  3.38 N*m

OR

0.29 kg*m² * α = 3.38 N*m

α = 11.655 rad/s²  ≈11.66 rad/s²

The angular acceleration is 11.66 rad/s²

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To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

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\omega = 359.03rad/s

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It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used
stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

a_{c} = rω²

ω² = a_{c} / r

ω = √( a_{c} / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a a_{c}  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

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A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

3 0
3 years ago
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