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andrew-mc [135]
2 years ago
9

A rod of length L has a mass density given by λ = λo(1 – x/L). What is the rod’s rotational inertia measured about the end where

x = 0? Write your answer in terms of its total mass, M, and length, L.
Physics
1 answer:
zhannawk [14.2K]2 years ago
6 0

Answer:

I = \frac{\lambda_o L^3}{12}

Explanation:

Rotational inertia of the rod about its one end is given as

I = \int dm x^2

here we know that

dm = \lambda dx

so we will have

I = \int (\lambda dx) x^2

I = \int \lambda_o(1 - x/L) x^2 dx

so we have

I = \lambda_o(\frac{L^3}{3} - \frac{L^3}{4})

I = \frac{\lambda_o L^3}{12}

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7 0
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Tom and his little sister are enjoying an afternoon at the ice rink. they playfully place their hands together and push against
Westkost [7]
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".

So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>F_t = F_s
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</span>m_t a_t = m_s a_s
<span>And find Tom's acceleration:
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5 0
2 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
2 years ago
In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

6 0
2 years ago
Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located
yan [13]

Answer:43.33 cm mark

Explanation:

Given

mass 1 is located at the 10 cm mark with weight of 15 kg

mass 2 is located at 60 cm mark with weight of 30 kg

string should be attached between them to balance the system

so the distance between the the two masses is 50 cm

For system to be balance torque of both the weight must nullify each other

Let us suppose string is at a distance of x cm from 15 kg mass so 30 kg mass is at a distance of 50-x cm

Balancing torque

15\times x-30\times (50-x)

x=\frac{100}{3}=33.33

so string should be at a mark of 10+33.33=43.33 cm

7 0
2 years ago
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