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disa [49]
3 years ago
10

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Physics
1 answer:
Fantom [35]3 years ago
8 0

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

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The ball will drop 0.881 m by the time it reaches the catcher.

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The position of the ball at time "t" is described by the position vector "r":

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Where:

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v0x = initial horizontal velocity.

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y0 = initial vertical position.

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When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

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With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

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y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

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t = 3.8*10^-6 s, so that

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