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1 year ago

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for the primitive yo-yo in fig, to find the downward acceleration of the cylinder and the tension in the string. you can take th

e toy as solid cylinder and icm=1/2mr2 , where m is the mass of the yo-yo toy

1 answer:

mixas84 [53]1 year ago

7 0

The downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

<h3>Downward acceleration of the cylinder</h3>

The downward acceleration of the solid cylinder is determined from the principle of conservation of angular momentum as shown below;

Iα = Tr

where;

I is moment of inertia of the solid cylinder
α is angular acceleration of the cylinder
T is tension in the string
r is length of the string

α = Tr/I

$\frac{a}{R} = \frac{Tr}{I} \\\\\frac{a}{R} = \frac{Tr}{\frac{1}{2} MR^2}\\\\\frac{a}{R} =\frac{2Tr}{MR^2} \\\\a = \frac{2Tr}{MR}$

where;

a is the downward acceleration of the solid cylinder
R is radius of the cylinder

Thus, the downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

Learn more about acceleration here: brainly.com/question/605631

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To compute the accelerations, we must find the function for a as the derivative of v

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$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

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$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

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The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

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