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1
The formula for the moment of inertia of an arbitrary body is written as \[ I = \sum m_i r_i^2 \] or \[ I = \int r^2 \, dm. \] |
0.25 |
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2
It is suggested to consider the square as a sum of rods, or alternatively, to introduce a surface mass density for integration. |
0.25 |
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3
The moment of inertia is obtained as \[ \dfrac{2 m a^2}{3}. \] |
0.50 |
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1
The relationship between the angular velocity of the wheels and the velocity of the car is written as \[ v = \omega \cdot r. \] |
0.10 |
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2
The angular velocity of the wheels at the top is obtained as \[ \omega_T = \dfrac{v_0}{a}. \] |
0.10 |
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3
Parallel axis theorem: \[ E_{\text{kin}} = E_{\text{translate}} + E_{\text{rotate}} \] |
0.10 |
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4
The kinetic energy of the car is obtained as \[ E_{\text{kin}} = \dfrac{3M + 10m}{6}\, v_0^2. \] |
0.20 |
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1
The distance from the wheel axle to the point of contact with the road at the “valley” is determined as \[ r = \sqrt{2}\, a. \] |
0.20 |
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2
An expression for the total kinetic energy of the car in terms of the angular velocity of the wheels and the distance from the axle to the road is obtained: \[ E_{\text{kin}} = \dfrac{1}{2} \left[ (M + 2m) r^2 \, \omega^2 + I \, \omega^2 \right]. \] |
0.40 |
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3
It is stated that the kinetic energy of the car at the “valley” of the road is equal to the kinetic energy of the car at the top: \[ E_{\text{kin, valley}} = E_{\text{kin, peak}}. \] |
0.20 |
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4
The angular velocity of the car at the “valley” of the road is determined as \[ \omega_V = \dfrac{v_0}{a} \sqrt{\dfrac{3M + 10m}{6M + 16m}}. \] |
0.20 |
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1
It is stated that the distance $G' O'$ remains constant and is equal to the distance $GO$: \[ G' O' = GO. \] |
0.20 |
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2
It is written that \[ G' O' = G' T' + T' O'. \] |
0.20 |
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3
It is written that \[ G' T' = \dfrac{a}{\cos \alpha}. \] |
0.20 |
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| 4 It is written that \[ T^\prime O^\prime = y(x)\] | 0.20 |
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| 5 Answer is obtained \[ \sqrt{2}a = \dfrac{a}{\cos \alpha} + y(x) \] | 0.20 |
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1
It is written that \[ \tan \alpha = -y^\prime(x),\] |
0.30 |
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2
The equation for the shape of the road is differentiated: \[ y'(x) = -\dfrac{h}{2a} \left( e^{x/a} - e^{-x/a} \right). \] |
0.40 |
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3
Answer is obtained \[ \tan \alpha = \dfrac{h}{2a} (e^{x/a}-e^{-x/a})\] |
0.30 |
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| 1 It is written that \[ \cos \alpha = \dfrac{1}{\sqrt{1+\tan^2 \alpha}}\] | 0.30 |
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2
The expression for the tangent of the inclination angle is substituted into the formula for the shape of the road: \[ \sqrt{2}a = a \sqrt{1 + \left( \dfrac{h}{2a} (e^{x/a} - e^{-x/a}) \right)^2} + k - h \cdot \dfrac{e^{x/a} + e^{-x/a}}{2}. \] |
0.30 |
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3
The case $x=0$ is considered, and the relationship between $k$ and $h$ is obtained: \[ \dfrac{k - h}{a} = \sqrt{2} - 1. \] |
0.50 |
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4
One of the two variables, $k$ or $h$, is expressed in terms of the other and substituted into the general equation. |
0.40 |
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5
The solution is obtained as \[ k = \sqrt{2}a. \] |
0.25 |
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6
The solution is obtained as \[ h = a. \] |
0.25 |
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1
It is stated that at the “valleys” of the road, $y = 0$. |
0.10 |
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2
The equation for $x$ at $y=0$ is written. |
0.10 |
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3
The equation is solved, and the positive root is obtained: \[ x_d = a \ln(\sqrt{2} + 1). \] |
0.10 |
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4
It is stated that \[ x_s = -x_d = -a \ln(\sqrt{2} + 1). \] |
0.10 |
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5
The solution for the length of the bump is obtained: \[ d = x_d - x_s = 2a \ln(\sqrt{2} + 1). \] |
0.10 |
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1
The equation for the shape of the road is substituted into the expression for the kinetic energy in terms of $r$: \[ r(x) = \sqrt{2}a - y(x). \] |
0.50 |
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2
The equation is solved, and the solution for $\omega(x)$ is obtained. |
1.00 |
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1
The equation is solved, and the solution for $v(x)$ is obtained. |
1.50 |
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