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3 years ago

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A solid cylinder of mass m and radius R has a string wound around it. A person holding the string pulls it vertically upward suc

h that the cylinder is suspended in midair for a brief time interval (change in)t and its center of mass does not move. The tension in the string is T, and the rotational inertia of the cylinder about its axis is (1/2)mR2. What is the linear acceleration of the person's hand during the time interval change in t?

1 answer:

LekaFEV [45]3 years ago

3 0

Answer:

α = 2T/mR

Explanation:

We know torque τ = Iα = TR where I = rotational inertia and α = linear (tangential) acceleration, T = tension in string and R = radius of cylinder.

Now T = tension in string and I = (1/2)mR²

So τ = Iα = TR

α = TR/I

substituting the values of the variables, we have

α = TR/(1/2)mR²

α = 2TR/mR²

α = 2T/mR

So, the linear acceleration of the person's hand on time, t is α = 2T/mR

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We are told that moment of inertia of System A consists of two of the larger disks. Thus;

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Where n is the number of smaller disks.

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Explanation:

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$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

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<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

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$h'=u_y'.t'+\frac{1}{2} g.t'^2$

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$t'=2.09\ s$

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$t_t=t'+t$

$t_t=2.09+2.041$

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