Five identical cylinders are each acted on by forces of equal magnitude. Which force exerts the biggest torque about the central
axes of the cylinders
1 answer:
Answer:
<em>From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.</em>
Explanation:
The image is shown below.
Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.
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Refer to the diagram shown.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.