Physics edexcel ial unit 4/5 discussion (15 and 21 june 2017)

Original post by Gintama-kun

i kinda think using a resultant force which is at an angle would nt explain the reason fully coz it will be havin a downward component ....

Original post by Gintama-kun

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jzt check whether da one i wrote is fine fr part c)
when amplitude increases the total energy of system increases At the bottom tot energy =KE so as KE increases ....

I agree with this part.

Original post by Gintama-kun

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Velocity increases causing upward centripetal force on girl to increase so at a point when centripetal force > than downward weight of gurl the girl will loose contact ....

I disagree (I will explain together with another point below that I disagree)

Original post by Gintama-kun

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As at the lowest point the resultant vertical force will be upwards so this guarantees her loosin contact ryt?

I disagree.

t seems to me that you confused resultant force with real forces. Centripetal force is a resultant force of the normal force and weight of the child.

When you said “Velocity increases causing upward centripetal force on girl to increase so at a point when centripetal force > than downward weight of the girl will loose contact”, you are treating centripetal force as a real force which is incorrect.

Even if the centripetal force is greater than the weight of the child, the child would not lose contact at the bottom of the swing. This is because the normal force (which is the contact force) at the bottom of the swing will be much greater than the weight in order to have a resultant force (the centripetal force) that is greater than the weight.

Consider a high-powered lift accelerating upward and you are inside the lift. Even if the resultant force is greater than your weight, you will not be caused to lift off the floor of the lift.

“As at the lowest point the resultant vertical force will be upwards so this guarantees her loosin contact ryt? ”
You need to realize that the resultant vertical force at the bottom of the swing is to cause the child to change the direction of motion of the child such that she can move in a “circular” motion.

Reply 187

Original post by Eimmanuel

I change a word.

😳 Why diju do that?! For half a minute there i thot i actually wrote smart 😳
Anyway.. U said the reaction is small but u dint explain why exactly it shud be small

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(edited 6 years ago)

Reply 188

Roe1777

Guys do we hv a separate topic for chem or do we discuss here?

Reply 189

Original post by Roe1777

Guys do we hv a separate topic for chem or do we discuss here?

Lol ofcourse not:
http://www.thestudentroom.co.uk/showthread.php?t=4752270

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Reply 190

Roe1777

Original post by shady zax

Lol ofcourse not:
http://www.thestudentroom.co.uk/showthread.php?t=4752270

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Thanks a bunch ^^

Reply 191

Gintama-kun

Original post by Eimmanuel

I hope you read the last statement a few times of post #158 (or is just below).

This resultant force can cause the child to lose touch with the seat because the normal force will be much smaller than the weight or is close to zero.

The weight could be most likely to be the resultant force and there is no other force to prevent the child from slipping off her seat.

I agree with this part.

I disagree (I will explain together with another point below that I disagree)

I disagree.

t seems to me that you confused resultant force with real forces. Centripetal force is a resultant force of the normal force and weight of the child.

When you said “Velocity increases causing upward centripetal force on girl to increase so at a point when centripetal force > than downward weight of the girl will loose contact”, you are treating centripetal force as a real force which is incorrect.

Even if the centripetal force is greater than the weight of the child, the child would not lose contact at the bottom of the swing. This is because the normal force (which is the contact force) at the bottom of the swing will be much greater than the weight in order to have a resultant force (the centripetal force) that is greater than the weight.

Consider a high-powered lift accelerating upward and you are inside the lift. Even if the resultant force is greater than your weight, you will not be caused to lift off the floor of the lift.

“As at the lowest point the resultant vertical force will be upwards so this guarantees her loosin contact ryt? ”
You need to realize that the resultant vertical force at the bottom of the swing is to cause the child to change the direction of motion of the child such that she can move in a “circular” motion.
ohh looks like i have mixed **** up sryy
kinda tried to relate this question to a car goin over a bump which looses contact at a velocity bt still its da lowest point ryt where contact is lost?

Reply 192

Original post by shady zax

😳 Why diju do that?! For half a minute there i thot i actually wrote smart 😳
Anyway.. U said the reaction is small but u dint explain why exactly it shud be small

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In fact, I may have to correct the statement.

The way that normal force is computed is depending on the motion of the child. Say the bottom of the swing,

N = mg + m\dfrac{v^2}{r}

When the swing is horizontal (at the max amplitude), the speed is zero and the weight is vertical, so the normal force is zero.

Reply 193

Original post by Eimmanuel

In fact, I may have to correct the statement.

The way that normal force is computed is depending on the motion of the child. Say the bottom of the swing,

N = mg + m\dfrac{v^2}{r}

When the swing is horizontal (at the max amplitude), the speed is zero and the weight is vertical, so the normal force is zero.

Ummm what about wen the swing is at an angle to the vertical?

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Reply 194

Original post by shady zax

Ummm what about wen the swing is at an angle to the vertical?

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N = mg \cos \theta + m\dfrac{v^2}{r}

where θ is the angle measured from the swing to the vertical line.

Reply 195

Gintama-kun

Original post by Eimmanuel

In fact, I may have to correct the statement.

The way that normal force is computed is depending on the motion of the child. Say the bottom of the swing,

N = mg + m\dfrac{v^2}{r}

When the swing is horizontal (at the max amplitude), the speed is zero and the weight is vertical, so the normal force is zero.

so at amplitude can it loose contact with swing coz normal reaction is zero

Reply 196

Original post by Eimmanuel

N = mg \cos \theta + m\dfrac{v^2}{r}

where θ is the angle measured from the swing to the vertical line.

Okay so.. thats making the reaction force vertical @_@

Dint u say its horizontal at max amplitude?

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Reply 197

Original post by Gintama-kun

so at amplitude can it loose contact with swing coz normal reaction is zero

Yes

Reply 198