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solniwko [45]
4 years ago
11

A solid disk has a total kinetic energy k. what is its rotational kinetic energy krot if it's rolling without slipping? express

your answer in terms of the variable k.
Physics
1 answer:
Dominik [7]4 years ago
5 0

the total energy is the sum of the linear and rotational energy: <span>
K = K_rot + K_lin 

first, we find the rotational kinetic energy of a rotating disc with an angular velocity of w. see the references for the moment of inertia of a disc. 
K_rot = (1/2)(I)(w^2) 
I = (1/2)(m)(r^2) 
K_rot = (1/4)(m)(r^2)(w^2) 

next, we find the linear kinetic energy of a rolling disc: 
K_lin = (1/2)(m)(v^2) 
v = angular velocity * circumference 
= w * (pi * 2 * r) 
K_lin = (1/2)(m)(w*2*pi*r)^2 
= (2*pi^2)(m)(r^2)(w^2) 

we find the total kinetic energy: 
K = K_rot + K_lin 
= (1/4)(m)(r^2)(w^2) + (2*pi^2)(m)(r^2)(w^2) 

and find the rotational contribution: 
K_rot = K * [K_rot/K] 
K_rot = K * [K_rot/(K_rot+K_lin)] 
K_rot = K * (1/4) / [(1/4) + (2*pi^2)] 
</span>K_rot = K / (8*pi^2 + 1)

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How did the angular acceleration change with the new moment of inertia? was your prediction correct?
elena55 [62]

In Newtonian physics, the acceleration of a body is inversely proportional to mass. In Newtonian rotational physics, angular acceleration is inversely proportional to the moment of inertia of a frame.

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6 0
1 year ago
An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst
sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

T_1 = 500°C

T_2 = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to T_{avg} = 535.5K \approx 550K

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

P_r = 0.63

A)

Number for the first strips is equal to

R_e_x = \frac{u_o.L}{v}

R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4

Calculating heat transfer coefficient from the first strip

h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3

h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2

The rate of convection heat transfer from the first strip is

q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W

The rate of convection heat transfer from the fifth trip is equal to

q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)

h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2

Calculating h_o_-_4

h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2

The rate of convection heat transfer from the tenth strip is

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)

h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2

Calculating

h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W

The rate of convection heat transfer from 25th strip is equal to

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)

Calculating h_o_-_2_5

h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2

Calculating h_o_-_2_4

h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W

6 0
4 years ago
A particle is projected with velocity v0 directly up a slope which makes an angle α with the horizontal. Assume frictionless mot
Nikolay [14]

Answer:

a)T total = 2*Voy/(g*sin( α ))

b)α = 0º ,  T total≅∞ (the particle, goes away horizontally indefinitely)

α = 90º,  T total=2*Voy/g

Explanation:

  • Velocity in the Y axis:

Voy=Vo*sinα

  • Time to reach the maximal height :

Kinematics equation: Vfy=Voy-at

a=g*sinα ;  g is gravity

if Vfy=0 ⇒ t=T ; time to reach the maximal height

so:

0=Voy-g*sin( α )*T

T=Voy/(g*sin( α ))

  • Time required to return to the starting point:

After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.

So T total= 2T = 2*Voy/(g*sin( α ))

  • α = 0º , sinα=0

The particle goes totally horizontal, goes away indefinitely

T total= 2*Voy/(g*sin( α )) ≅∞

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T total=2*Voy/g

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