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Is it wrong to imagine the gravitational potential of a system based on the center of mass?
Clash Royale CLAN TAG#URR8PPP
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I was practising some questions until I stumbled upon a question. It is as shown in the picture below:
I am only concerned with part (a) of this question which is trivial to solve. The only thing which I don't understand is why I'm getting different answers when solving through different methods.
Method 1: (Integration)
The problem can be reduced to, $$V=int_0^l/rfracmlgRcdot Rcostheta,dtheta=dfracmR^2glsinbiggl(dfraclRbiggl)$$
which is stated as the answer in the book.
Method 2: (Center of Mass)
Assuming that the whole mass of the system can be taken on the center of mass, therefore, the center of mass of the chain will subtend an angle $alpha=dfracl2R$ because $alpha propto arc$, i.e $dfracl2$.
$$implies V=mgcdot Rcosbiggl(fracl2Rbiggl)$$
To summarize myself, I'm curious as to why the second method doesn't yield the expected answer? Is it wrong to calculate the gravitational potential energy of a body through its center of mass?
newtonian-mechanics newtonian-gravity work potential-energy
edited 1 hour ago
Qmechanic♦
98.6k121781085
asked 3 hours ago
Utkarsh Verma
206
add a comment |Â
up vote
2
down vote
favorite
I was practising some questions until I stumbled upon a question. It is as shown in the picture below:
I am only concerned with part (a) of this question which is trivial to solve. The only thing which I don't understand is why I'm getting different answers when solving through different methods.
Method 1: (Integration)
The problem can be reduced to, $$V=int_0^l/rfracmlgRcdot Rcostheta,dtheta=dfracmR^2glsinbiggl(dfraclRbiggl)$$
which is stated as the answer in the book.
Method 2: (Center of Mass)
Assuming that the whole mass of the system can be taken on the center of mass, therefore, the center of mass of the chain will subtend an angle $alpha=dfracl2R$ because $alpha propto arc$, i.e $dfracl2$.
$$implies V=mgcdot Rcosbiggl(fracl2Rbiggl)$$
To summarize myself, I'm curious as to why the second method doesn't yield the expected answer? Is it wrong to calculate the gravitational potential energy of a body through its center of mass?
newtonian-mechanics newtonian-gravity work potential-energy
edited 1 hour ago
Qmechanic♦
98.6k121781085
asked 3 hours ago
Utkarsh Verma
206
Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
– Chair
1 hour ago
Yes, I also had that in mind initially, but I was mistaken that this question had a figure, so I posted a picture but forgot about it in the process. I will make required edits soon.
– Utkarsh Verma
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was practising some questions until I stumbled upon a question. It is as shown in the picture below:
I am only concerned with part (a) of this question which is trivial to solve. The only thing which I don't understand is why I'm getting different answers when solving through different methods.
Method 1: (Integration)
The problem can be reduced to, $$V=int_0^l/rfracmlgRcdot Rcostheta,dtheta=dfracmR^2glsinbiggl(dfraclRbiggl)$$
which is stated as the answer in the book.
Method 2: (Center of Mass)
Assuming that the whole mass of the system can be taken on the center of mass, therefore, the center of mass of the chain will subtend an angle $alpha=dfracl2R$ because $alpha propto arc$, i.e $dfracl2$.
$$implies V=mgcdot Rcosbiggl(fracl2Rbiggl)$$
To summarize myself, I'm curious as to why the second method doesn't yield the expected answer? Is it wrong to calculate the gravitational potential energy of a body through its center of mass?
newtonian-mechanics newtonian-gravity work potential-energy
edited 1 hour ago
Qmechanic♦
98.6k121781085
asked 3 hours ago
Utkarsh Verma
206
I was practising some questions until I stumbled upon a question. It is as shown in the picture below:
I am only concerned with part (a) of this question which is trivial to solve. The only thing which I don't understand is why I'm getting different answers when solving through different methods.
Method 1: (Integration)
The problem can be reduced to, $$V=int_0^l/rfracmlgRcdot Rcostheta,dtheta=dfracmR^2glsinbiggl(dfraclRbiggl)$$
which is stated as the answer in the book.
Method 2: (Center of Mass)
Assuming that the whole mass of the system can be taken on the center of mass, therefore, the center of mass of the chain will subtend an angle $alpha=dfracl2R$ because $alpha propto arc$, i.e $dfracl2$.
$$implies V=mgcdot Rcosbiggl(fracl2Rbiggl)$$
To summarize myself, I'm curious as to why the second method doesn't yield the expected answer? Is it wrong to calculate the gravitational potential energy of a body through its center of mass?
newtonian-mechanics newtonian-gravity work potential-energy
newtonian-mechanics newtonian-gravity work potential-energy
edited 1 hour ago
Qmechanic♦
98.6k121781085
asked 3 hours ago
Utkarsh Verma
206
edited 1 hour ago
Qmechanic♦
98.6k121781085
asked 3 hours ago
Utkarsh Verma
206
edited 1 hour ago
Qmechanic♦
98.6k121781085
edited 1 hour ago
Qmechanic♦
98.6k121781085
edited 1 hour ago
Qmechanic♦
98.6k121781085
98.6k121781085
asked 3 hours ago
Utkarsh Verma
206
asked 3 hours ago
Utkarsh Verma
206
asked 3 hours ago
Utkarsh Verma
206
206
Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
– Chair
1 hour ago
Yes, I also had that in mind initially, but I was mistaken that this question had a figure, so I posted a picture but forgot about it in the process. I will make required edits soon.
– Utkarsh Verma
1 hour ago
add a comment |Â
Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
– Chair
1 hour ago
Yes, I also had that in mind initially, but I was mistaken that this question had a figure, so I posted a picture but forgot about it in the process. I will make required edits soon.
– Utkarsh Verma
1 hour ago
Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
– Chair
1 hour ago
Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
– Chair
1 hour ago
Yes, I also had that in mind initially, but I was mistaken that this question had a figure, so I posted a picture but forgot about it in the process. I will make required edits soon.
– Utkarsh Verma
1 hour ago
Yes, I also had that in mind initially, but I was mistaken that this question had a figure, so I posted a picture but forgot about it in the process. I will make required edits soon.
– Utkarsh Verma
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You are assuming that the COM lies on the chain. This is not the case. The COM will be closer to the center of the sphere than the chain. The COM of an extended body need not be within the body itself.
Let's look at the height of COM coordinate.
$$my_com=int y dm=lambdaint y ds=lambdaint_0^l/R Rcosalphacdot R dalpha$$
This is your first integral (without the $mg$ term). Therefore the methods are equivalent.
For a general object of mass $M$
$$Mgy_com=int gy dm=int dU=U$$
So the two integrals are in fact equal in general, assuming $g$ is uniform. If $g$ is not uniform then you just use the center of gravity instead, but the arguments still play out the same.
answered 3 hours ago
Aaron Stevens
5,8462830
1
Ah, that was a huge blunder on my part. Thanks for the quick reply.
– Utkarsh Verma
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You are assuming that the COM lies on the chain. This is not the case. The COM will be closer to the center of the sphere than the chain. The COM of an extended body need not be within the body itself.
Let's look at the height of COM coordinate.
$$my_com=int y dm=lambdaint y ds=lambdaint_0^l/R Rcosalphacdot R dalpha$$
This is your first integral (without the $mg$ term). Therefore the methods are equivalent.
For a general object of mass $M$
$$Mgy_com=int gy dm=int dU=U$$
So the two integrals are in fact equal in general, assuming $g$ is uniform. If $g$ is not uniform then you just use the center of gravity instead, but the arguments still play out the same.
answered 3 hours ago
Aaron Stevens
5,8462830
1
Ah, that was a huge blunder on my part. Thanks for the quick reply.
– Utkarsh Verma
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
You are assuming that the COM lies on the chain. This is not the case. The COM will be closer to the center of the sphere than the chain. The COM of an extended body need not be within the body itself.
Let's look at the height of COM coordinate.
$$my_com=int y dm=lambdaint y ds=lambdaint_0^l/R Rcosalphacdot R dalpha$$
This is your first integral (without the $mg$ term). Therefore the methods are equivalent.
For a general object of mass $M$
$$Mgy_com=int gy dm=int dU=U$$
So the two integrals are in fact equal in general, assuming $g$ is uniform. If $g$ is not uniform then you just use the center of gravity instead, but the arguments still play out the same.
answered 3 hours ago
Aaron Stevens
5,8462830
1
Ah, that was a huge blunder on my part. Thanks for the quick reply.
– Utkarsh Verma
3 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You are assuming that the COM lies on the chain. This is not the case. The COM will be closer to the center of the sphere than the chain. The COM of an extended body need not be within the body itself.
Let's look at the height of COM coordinate.
$$my_com=int y dm=lambdaint y ds=lambdaint_0^l/R Rcosalphacdot R dalpha$$
This is your first integral (without the $mg$ term). Therefore the methods are equivalent.
For a general object of mass $M$
$$Mgy_com=int gy dm=int dU=U$$
So the two integrals are in fact equal in general, assuming $g$ is uniform. If $g$ is not uniform then you just use the center of gravity instead, but the arguments still play out the same.
answered 3 hours ago
Aaron Stevens
5,8462830
You are assuming that the COM lies on the chain. This is not the case. The COM will be closer to the center of the sphere than the chain. The COM of an extended body need not be within the body itself.
Let's look at the height of COM coordinate.
$$my_com=int y dm=lambdaint y ds=lambdaint_0^l/R Rcosalphacdot R dalpha$$
This is your first integral (without the $mg$ term). Therefore the methods are equivalent.
For a general object of mass $M$
$$Mgy_com=int gy dm=int dU=U$$
So the two integrals are in fact equal in general, assuming $g$ is uniform. If $g$ is not uniform then you just use the center of gravity instead, but the arguments still play out the same.
answered 3 hours ago
Aaron Stevens
5,8462830
edited 2 hours ago
answered 3 hours ago
Aaron Stevens
5,8462830
answered 3 hours ago
Aaron Stevens
5,8462830
answered 3 hours ago
Aaron Stevens
5,8462830
5,8462830
1
Ah, that was a huge blunder on my part. Thanks for the quick reply.
– Utkarsh Verma
3 hours ago
add a comment |Â
1
Ah, that was a huge blunder on my part. Thanks for the quick reply.
– Utkarsh Verma
3 hours ago
1
1
Ah, that was a huge blunder on my part. Thanks for the quick reply.
– Utkarsh Verma
3 hours ago
Ah, that was a huge blunder on my part. Thanks for the quick reply.
– Utkarsh Verma
3 hours ago
add a comment |Â
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Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
– Chair
1 hour ago
Yes, I also had that in mind initially, but I was mistaken that this question had a figure, so I posted a picture but forgot about it in the process. I will make required edits soon.
– Utkarsh Verma
1 hour ago