Inertia and Torque
http://filestore.aqa.org.uk/subjects/AQA-PHYA5-2C-QP-JUN12.PDF
In question 2, I got:
a) F x r = 32 x 0.15 = 4.8
a)ii) Output power = Indicated power - Friction power
2600x(2pi/60) x (4.8-1.2)
= 980 W
Why do they add the torques? I'm so confused
Can someone please help?
In question 2, I got:
a) F x r = 32 x 0.15 = 4.8
a)ii) Output power = Indicated power - Friction power
2600x(2pi/60) x (4.8-1.2)
= 980 W
Why do they add the torques? I'm so confused
Can someone please help?
(edited 8 years ago)
Yeah sorry, I deleted my post because I realised I misinterpreted "output power". I assume this means the power due to energy transfer from the machine to the cutting of the wood. For that, you want the equation using a torque of 4.8 Nm only, i.e.
Output power = 2600x(2pi/60) x (4.8 Nm)
The rest of the power loss is dissipated as heat internally (due to the torque of 1.2 Nm). The power drawn from the plug socket is the sum of these.
Output power = 2600x(2pi/60) x (4.8 Nm)
The rest of the power loss is dissipated as heat internally (due to the torque of 1.2 Nm). The power drawn from the plug socket is the sum of these.
(edited 8 years ago)
This one looks a bit ambiguous. If a fraction of the power is dissipated internally in the bearings of the motor then it should not strictly contribute to the "output power".
I would hope that calculating total electrical power in (using 6 Nm) and useful "cutting" power out (using 4.8 Nm) would be enough to cover your bases.
I would hope that calculating total electrical power in (using 6 Nm) and useful "cutting" power out (using 4.8 Nm) would be enough to cover your bases.
No problem!
Each torque is a source of power loss. The 4.8 Nm torque due to the wood will give a power loss of 2600x(2pi/60)x(4.8), which is the power devoted to actually cutting the wood. The 1.2 Nm torque due to internal friction will give a power loss of 2600x(2pi/60)x(1.2), which is the power lost to that friction. Maybe that energy causes the motor to heat up or make noise.
The total power that is expended by the motor in cutting and friction is the sum of these, 2600x(2pi/60)x(4.8) + 2600x(2pi/60)x(1.2). You can arrive at this number by adding the torques in advance, 2600x(2pi/60)x(4.8+1.2). It's the same number.
Due to conservation of energy, that is the amount of power that the motor will draw from the plug socket.
Each torque is a source of power loss. The 4.8 Nm torque due to the wood will give a power loss of 2600x(2pi/60)x(4.8), which is the power devoted to actually cutting the wood. The 1.2 Nm torque due to internal friction will give a power loss of 2600x(2pi/60)x(1.2), which is the power lost to that friction. Maybe that energy causes the motor to heat up or make noise.
The total power that is expended by the motor in cutting and friction is the sum of these, 2600x(2pi/60)x(4.8) + 2600x(2pi/60)x(1.2). You can arrive at this number by adding the torques in advance, 2600x(2pi/60)x(4.8+1.2). It's the same number.
Due to conservation of energy, that is the amount of power that the motor will draw from the plug socket.
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