A level physics - Engineering question (5)
The fly-press shown below is used by a jeweller to punch shapes out of a thin metal sheet.
The frame holds a screw and punch. Two arms are attached to the screw, each loaded with a heavy steel ball. The screw is driven downwards when the arms are rotated.
Kinetic energy is stored in the rotating parts: the balls, arms, screw and punch. This energy is used to punch the shape out of the metal sheet.
Q)For thicker or stiffer metal sheets the rotational kinetic energy at 2.9 rev s–1 is not enough to punch out the shape.
The distance from the axis of rotation to the centre of each ball is y.
The radius of each ball is R.
The stored energy can be increased by
either
• increasing y by 15% without changing R
or
• increasing R by 15% without changing y.
Deduce which of these would produce the greater increase in stored energy.
- I got to the right conclusion that increasing R by 15% without changing y would produce the greater increase in stored energy I just used the logic of I=mr^2 and r=R+y so is R is bigger increasing R would obviously do more.
-However the mark scheme says:
(I = 2 mr2 and Ek = ½ I ω2)
Increasing y by 15% gives new I = 1.15^2 × original I (or 1.32) ✔
Increasing R by 15% increases I by 1.15^3 (or 1.52) ✔
Second option gives greater increase in I, and Ek
also increased (by same ratio). ✔
Accept answers without calculation:
I prop to y^2 ✔
I prop to R^3 ✔
I don't understand how they got
>I prop to y^2 ✔
>I prop to R^3 ✔
The frame holds a screw and punch. Two arms are attached to the screw, each loaded with a heavy steel ball. The screw is driven downwards when the arms are rotated.
Kinetic energy is stored in the rotating parts: the balls, arms, screw and punch. This energy is used to punch the shape out of the metal sheet.
Q)For thicker or stiffer metal sheets the rotational kinetic energy at 2.9 rev s–1 is not enough to punch out the shape.
The distance from the axis of rotation to the centre of each ball is y.
The radius of each ball is R.
The stored energy can be increased by
either
• increasing y by 15% without changing R
or
• increasing R by 15% without changing y.
Deduce which of these would produce the greater increase in stored energy.
- I got to the right conclusion that increasing R by 15% without changing y would produce the greater increase in stored energy I just used the logic of I=mr^2 and r=R+y so is R is bigger increasing R would obviously do more.
-However the mark scheme says:
(I = 2 mr2 and Ek = ½ I ω2)
Increasing y by 15% gives new I = 1.15^2 × original I (or 1.32) ✔
Increasing R by 15% increases I by 1.15^3 (or 1.52) ✔
Second option gives greater increase in I, and Ek
also increased (by same ratio). ✔
Accept answers without calculation:
I prop to y^2 ✔
I prop to R^3 ✔
I don't understand how they got
>I prop to y^2 ✔
>I prop to R^3 ✔
(edited 2 months ago)
Original post by 1234kelly
The fly-press shown below is used by a jeweller to punch shapes out of a thin metal sheet.
...
...
Shown where?
Original post by Eimmanuel
Shown where?
Original post by 1234kelly
The fly-press shown below is used by a jeweller to punch shapes out of a thin metal sheet.
The frame holds a screw and punch. Two arms are attached to the screw, each loaded with a heavy steel ball. The screw is driven downwards when the arms are rotated.
Kinetic energy is stored in the rotating parts: the balls, arms, screw and punch. This energy is used to punch the shape out of the metal sheet.
Q)For thicker or stiffer metal sheets the rotational kinetic energy at 2.9 rev s–1 is not enough to punch out the shape.
The distance from the axis of rotation to the centre of each ball is y.
The radius of each ball is R.
The stored energy can be increased by
either
• increasing y by 15% without changing R
or
• increasing R by 15% without changing y.
Deduce which of these would produce the greater increase in stored energy.
- I got to the right conclusion that increasing R by 15% without changing y would produce the greater increase in stored energy I just used the logic of I=mr^2 and r=R+y so is R is bigger increasing R would obviously do more.
-However the mark scheme says:
(I = 2 mr2 and Ek = ½ I ω2)
Increasing y by 15% gives new I = 1.15^2 × original I (or 1.32) ✔
Increasing R by 15% increases I by 1.15^3 (or 1.52) ✔
Second option gives greater increase in I, and Ek
also increased (by same ratio). ✔
Accept answers without calculation:
I prop to y^2 ✔
I prop to R^3 ✔
I don't understand how they got
>I prop to y^2 ✔
>I prop to R^3 ✔
The frame holds a screw and punch. Two arms are attached to the screw, each loaded with a heavy steel ball. The screw is driven downwards when the arms are rotated.
Kinetic energy is stored in the rotating parts: the balls, arms, screw and punch. This energy is used to punch the shape out of the metal sheet.
Q)For thicker or stiffer metal sheets the rotational kinetic energy at 2.9 rev s–1 is not enough to punch out the shape.
The distance from the axis of rotation to the centre of each ball is y.
The radius of each ball is R.
The stored energy can be increased by
either
• increasing y by 15% without changing R
or
• increasing R by 15% without changing y.
Deduce which of these would produce the greater increase in stored energy.
- I got to the right conclusion that increasing R by 15% without changing y would produce the greater increase in stored energy I just used the logic of I=mr^2 and r=R+y so is R is bigger increasing R would obviously do more.
-However the mark scheme says:
(I = 2 mr2 and Ek = ½ I ω2)
Increasing y by 15% gives new I = 1.15^2 × original I (or 1.32) ✔
Increasing R by 15% increases I by 1.15^3 (or 1.52) ✔
Second option gives greater increase in I, and Ek
also increased (by same ratio). ✔
Accept answers without calculation:
I prop to y^2 ✔
I prop to R^3 ✔
I don't understand how they got
>I prop to y^2 ✔
>I prop to R^3 ✔
Moment of inertia is proportional to y^2 because they are just using I = mr^2 where r = y.
Next, the moment of inertia is proportional to R^3 is shown in the MS below. pay attention to the highlighted portion in yellow with a marking circled in red.
Post the whole MS in future and don’t assume the “trivial” info.
I disagree with their agreement if you study moment of inertia at University, you would know why. I guess they are simplifying the calculation to suit A level physics.
Original post by Eimmanuel
Moment of inertia is proportional to y^2 because they are just using I = mr^2 where r = y.
Next, the moment of inertia is proportional to R^3 is shown in the MS below. pay attention to the highlighted portion in yellow with a marking circled in red.
Post the whole MS in future and don’t assume the “trivial” info.
I disagree with their agreement if you study moment of inertia at University, you would know why. I guess they are simplifying the calculation to suit A level physics.
Next, the moment of inertia is proportional to R^3 is shown in the MS below. pay attention to the highlighted portion in yellow with a marking circled in red.
Post the whole MS in future and don’t assume the “trivial” info.
I disagree with their agreement if you study moment of inertia at University, you would know why. I guess they are simplifying the calculation to suit A level physics.
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