Rotational dynamics of cage

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A mine cage of mass 4 tonne is to be raised with an acceleration of 1.5m/s^2
using a hoist drum of 1.5m diameter. The drum’s mass is 750kg and its radius of gyration is 600mm.The effect of bearing friction is equivalent to a torque of 3kNm at the hoist drum. What isthe driving torque required on the drum ? If the driving torque ceases when the load is
moving upwards at 6m/s, find the deceleration of the load and how far it travels before
coming to rest.

My question is the second part, to find the deceleration.

i already know the answer to the first part, which is the torqure required, which is 37440Nm

I get F = M*A >> A = F/M >>> 37440/4000 = 9.36m/s^2

but i have the answer to be 9.64m/s^2
 

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  • #2
haruspex
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the answer to the first part, which is the torqure required, which is 37440Nm
Typo? I make 37470.
I get F = M*A >> A = F/M >>> 37440/4000 = 9.36m/s^2
What does dividing a torque by a mass give you?
Draw the free body diagrams, putting in unknowns for tension and acceleration. Develop the equations for the cage and the drum separately.
(It doesn't happen here, but in general you ought to check the tension has not gone negative. It could with greater drum friction.)
 
  • #3
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Going back to the first question, could you tell me the error in my calculation to work out driving torque

mass*acceleration*radius + mass*gravity*radius + inertia * radius of gyration2 + friciton torque = driving torque

4000*1.5*0.75 + 4000*9.81*0.75 + 750*0.62 + 3000 = 37.2*103
 

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