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3 years ago

15

A 42.0-cm diameter disk rotates with a constant angular acceleration of 2.70 rad/s2. It starts from rest at t = 0, and a line dr

awn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.
(a) Find the angular speed of the wheel at t=2.30s.

(b) Find the linear velocity and tangential acceleration of P at t=2.30s.

(c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.

1 answer:

Zielflug [23.3K]3 years ago

7 0

Answer:

6.21 rad/s

1.3041 m/s, 0.567 m/s²

$106.4778\ ^{\circ}$

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration = 2.3 rad/s²

$\theta$ = Angle of rotation

t = Time taken = 2.3 s

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=0+2.7\times 2.3\\\Rightarrow \omega_f=6.21\ rad/s$

The angular speed is 6.21 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=0.21\times 6.21\\\Rightarrow v=1.3041\ m/s$

Linear velocity is 1.3041 m/s

Tangential acceleration is given by

$a_t=r\alpha\\\Rightarrow a_t=0.21\times 2.7\\\Rightarrow a_t=0.567\ m/s^2$

Tangential acceleration is 0.567 m/s²

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2.7\times 2.3^2\\\Rightarrow \theta=7.1415\ rad$

In degress the angle would be

$57.3+7.1415\times \dfrac{180}{\pi}=466.47780\ ^{\circ}$

From x axis it would be

$466.47780-360=106.4778\ ^{\circ}$

The angle is $106.4778\ ^{\circ}$ from x axis

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A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

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Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

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b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

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Work done, W = nRT ln(V₂/V₁)

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Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

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Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

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W₄ = 0 J

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W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

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Answer:

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Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

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$I_{k_1} = I_{p_1 } = m * r_i^2$

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$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

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=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

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