Rotation about a fixed axis problem

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SUMMARY

The discussion revolves around a physics problem involving rotation about a fixed axis, specifically analyzing a hoisting mechanism with two pulleys. The problem states that a constant force of 1.5 kN is applied to a system with pulleys of radii 0.4 m and 0.8 m, and a block A with a mass of 300 kg is attached. The participants calculated the moment of inertia (I) as 25 kg·m² and attempted to derive the angular acceleration (α) and tension (T) in the system. The final answers provided were α = 0.312 rad/s² (CW), T = 2980.5 N, and a vertical acceleration (a) of 0.125 m/s².

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  • Understanding of rotational dynamics and moment of inertia (I = mk²)
  • Knowledge of Newton's second law for rotation and linear motion
  • Familiarity with trigonometric functions, particularly sine for force components
  • Ability to interpret and manipulate equations of motion in a mechanical system
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  • Study the principles of rotational dynamics, focusing on torque and angular acceleration
  • Learn how to resolve forces into components, especially in inclined systems
  • Explore the concept of radius of gyration and its application in calculating moment of inertia
  • Investigate the effects of different pulley configurations on mechanical advantage and acceleration
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Rob K
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Hi guys I was trying not to post this as it is difficult to understand without the diagram, but I will try and give you a visual. Basically there are 2 pulleys attached together on the one pivot, the larger one is driven by a belt that is driven by a motor. The inner pulley is used to hoist the weight.

Homework Statement


Figure 3 shows a block A attached to a hoisting mechanism which consists of two
drums of radii r1 = 0:4 m, r2 = 0:8 m, ¯xed at O and which have a combined mass
of m1 = 100 kg and a radius of gyration about O of kO = 0:5 m. If a constant force
P = 1:5 kN is applied by the power unit, which consists of a pulley of radius r3
attached to a motor, determine the vertical acceleration of block A. The cord from
the motor to the drum is at an angle 40± to the horizontal. Block A has a mass
m2 = 300kg.


Homework Equations


I = mk2

+CCW \SigmaMo = Io\alpha

+\uparrow\SigmaFy = m(ag)x

+CCW a = \alphar

a = \alphar

The Attempt at a Solution


I = 100kg * (0.5m)^{2} = 25kg.m2

\SigmaMo
T(0.4m) = (25 kg.m2\alpha

+\uparrow\SigmaFy = m(ag)x
-300(9.81)N + 1500sin40N + T = - 300a

a = \alpha0.4

I worked out \alpha to be 5.56, or 24. something. Basically after 20 odd hours, I have given up, there is something clearly very wrong and missing, and between me and 2 other people we don't know where to go from here.

The answers are as follows
[\alpha = 0:312 rads/s2CW, T = 2980:5 N, a = 0:125 m/s2.]

Please help.

Kind Regards

Rob K
 
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Can anyone help with this?

Do I need to provide some more information?
 
Try to attach a picture. And explain your notations. What do you mean by ":"? Is it for decimal point? ehild
 
Last edited:

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