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madam [21]
3 years ago
9

A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical a

xle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Physics
1 answer:
VashaNatasha [74]3 years ago
8 0

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

Radius of the disk, r = 14.9 cm = 0.149 m

Rotational Inertia, I = 5.92*10^-3 kgm²

Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

b)

The mechanical energy of the cockroach is not converted as it stops

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<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)

The total number of ions present in the solution are 3.

  • To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (Ca(NO_3)_2) = 11.3 g

M_{solute} = Molar mass of solute (Ca(NO_3)_2) = 164  g/mol

W_{solvent} = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=iK_fm

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i = Vant hoff factor = 3

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m = molality of solution = 0.599 m

Putting values in above equation, we get:

\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC

Hence, the freezing point of solution is -3.34°C

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