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suter [353]
4 years ago
13

Challenge Problem (Extra Credit) Your good friend’s car (mentioned above) ignores a stop sign and enters an intersection with a

speed v0 of [05] 31.4 m/s. He then continues with a constant velocity. A motorcycle traffic patrol officer is parked at the intersection. He discards his hot cocoa and doughnut and begins accelerating (at the instant the car passes him) with a constant acceleration of 3.29 m/s2.(a) How long (s) does it take the officer to catch the car?(b) How far (m) has the officer traveled when he catches up to the car?(c) How fast (m/s) is the officer going when he catches up to the car?
Physics
1 answer:
GrogVix [38]4 years ago
6 0

Answer:

a) 19.1s

b) 599.4m

c) 62.8 m/s

Explanation:

Suppose both the officer and your good friend start at the same position (the intersection point). The equation of motion for each of them are:

- Your friend: s = v_0t = 31.4t m

- The Officer: s = at^2/2 = 3.29t^2/2 = 1.645 t^2 m

(a) In order for the officer to catch him, they should both end up in the same position at the same time (other than t = 0)

31.4t = 1.645t^2

t = 31.4/1.645 = 19.1 s

(b) Within 19.1 s, the officer would have travelled a distance of

s = 1.645*19.1^2 = 599.4 m

(c) His speed when he catches up would be

v = at = 3.29*19.1 = 62.8 m/s

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

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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

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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}

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\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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