Course - BTEC Level 4 Higher National Certificate in Engineering

Dynamic Mechanical Systems

You are required to submit a well written detailed explanation of the energy changes that occur for the following systems.

a) A block 50kg starting at rest sliding down a slope of 1 in 7 (tan) for 50m, with a dynamic friction coefficient of 0.023.

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Objective: To find initial and final energies and energy losses that may occur during the process.
Weight of the block is 50kg
The slope through which it is falling= 1 in 7.
Which means it is of angle 8.1340
The distance for which the block has moved down the slope= 50m
Dynamic friction Coefficient = 0.023
Suppose the distance moved is 50 metres
The actual height travelled will be 7.0711m (From the slope of 1 in 7)

Initial energy of the block will be due to the potential energy contained in the block by virtue of its position. Since the block is lying at a height of 7.0711 metre height, the potential energy contained in it will be,

PE = m*g*h = 50 * 7.0711 = 363.56 J

Since the body is on the slant, the actual normal force will be different from the actual reaction by the slope, Depending on the actual reaction from the floor and the dynamic friction force acting there on, some of the energy contained in the block will be consumed to overcome the friction force.

Since the work done in this case is 50.3125J the remaining energy(final energy) will be equal to 313.2475J

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Question: A 200kg cylinder rolling freely down a slope from rest of 1 in 30 (sine) for 20m. Indicate in your explanation why the cylinder rolls including whether friction is static of dynamic.

B) The actual cylinder rolling down the plane at an angle of 1 in 30 sine, for 20meters,

The reason for the cylinder to roll down the plane is the fact that there is net gravity force acting along the surface of the plane, which is making the cylinder roll down the plane(Meriam & Craige,2012).

In the current case, as the cylinder is rolling down the plane, there will be certain amount of friction acting between the cylinder and the plane, but the friction is not static rather it is dynamic in nature, it resists the relative motion between the two bodies.

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Question: The wheels each have a mass of A kg and radius of gyration of 300mm, the outer diameter of the wheel is 500mm.
a. Calculate the angular velocity of each wheel.
b. Calculate the moment of inertia of each wheel.
c. Calculate the magnitude of gyroscopic torques produced on the bike.
d. Using research explain the effect of gyroscopic torque on the motorbike.

A motor cycle and the rider travel around a right hand bend of radius 25m at 60kmph
The mass of each wheel is 17.4kg
The outer diameter of the wheel is 500mm

Outer diameter of the wheel is 500mm, so the angular velocity of the each wheel is,
Velocity V = r* (omega)

Angular velocity = [60*1000/(3600)]/0.25metres = 66.7 radians per second.

Moment of inertia of the each wheel = m*r2= 17.4*(0.25)2 = 1.0875 kg-m2.

Gyroscopic torquesproduced on the bike = 1.0875 x 66.7 x9.42 = 683.23 J
(On each wheel - Combine together it is going to be 1366.46 J)
Precision angular velocity = 2*3.14* (25)/(60000/3600) = 9.42

FAQ for Dynamic Mechanical Systems

• What are dynamic mechanical systems?
• Why are dynamic mechanical systems important?
• What are some real-world applications of studying dynamic mechanical systems?
• How do we analyze dynamic mechanical systems?
• What are some common challenges in analyzing dynamic mechanical systems?

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