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marta [7]
3 years ago
10

A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.A

t the instant when d = 30 cm of string has unwound off the disk, what is the torque exerted about the center of the disk?
Physics
1 answer:
kenny6666 [7]3 years ago
7 0

Answer: 1.76 Nm

Explanation:

If the force pulls horizontally, this means that the force is tangent to the disk at any point of the string unwinding process, so the distance d is irrelevant.

In this case, the torque is directly given by the product of the force times the distance perpendicular to the center of the disk, which is just the radius, as follows:

τ = F * r = 16 N. (0.11) m = 1.76 Nm

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If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
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Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

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3 years ago
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