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Effect of friction between pulley and th...

Effect of friction between pulley and thread :
In ideal cases i.e., when pulley and strings are massless and no friction exists at any contact surface, then tension in the string is constant throughout its length. But consider a, massless pulley and massless string but friction exists between pulley and string With coefficient `mu`. Then tension at the two ends of the pulley will be different. As shown in figure, consider an element of string :
`dN=(T+dT) "sin" (d theta)/2+T "sin"(d theta)/2`

`(T+dT)"cos"(d theta)/2 -T"cos"(d theta)/2-mu dN=dr a=0` (massless strings)
`dT"cos"(d theta)/2=mu dN`
`dT"cos"(d theta)/2=mu [(T+dt) "sin" (d theta)/2+T "sin" (d theta)/2]`
`dt."cos" (d theta)/2=mu[T "sin" (d theta)/2+dT. "sin" (d theta)/2+T "sin" (d theta)/2]`
`dT=mu[T. (d theta)/2+0+T(d theta)/2]=mu T cos theta`
int_(T_(1))^(T_(2)) (d T)/T=int _(0)^(pi)mu d theta rArr ln(T_(2)/T_(1))=mu pi rArrT_(2)/T_(1)=e^(mu pi)`
Suppose cofficient of friction between the string and pulley is `mu =1/pi`
What should be the ratio of heavier mass to lighter mass for no motion ?

A

`e`

B

`1/e`

C

`e^(L)`

D

`e^(R)`

Text Solution

Verified by Experts

The correct Answer is:
a
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Knowledge Check

  • Effect of friction between pulley and thread : In ideal cases i.e., when pulley and strings are massless and no friction exists at any contact surface, then tension in the string is constant throughout its length. But consider a, massless pulley and massless string but friction exists between pulley and string With coefficient mu . Then tension at the two ends of the pulley will be different. As shown in figure, consider an element of string : dN=(T+dT) "sin" (d theta)/2+T "sin"(d theta)/2 (T+dT)"cos"(d theta)/2 -T"cos"(d theta)/2-mu dN=dr a=0 (massless strings) dT"cos"(d theta)/2=mu dN dT"cos"(d theta)/2=mu [(T+dt) "sin" (d theta)/2+T "sin" (d theta)/2] dt."cos" (d theta)/2=mu[T "sin" (d theta)/2+dT. "sin" (d theta)/2+T "sin" (d theta)/2] dT=mu[T. (d theta)/2+0+T(d theta)/2]=mu T cos theta int_(T_(1))^(T_(2)) (d T)/T=int _(0)^(pi)mu d theta rArr ln(T_(2)/T_(1))=mu pi rArrT_(2)/T_(1)=e^(mu pi) Suppose cofficient of friction between the string and pulley is mu =1/pi The tension on side ofheavier\nass will be:

    A
    `m_(1)g`
    B
    `m_(2)g`
    C
    `(2m_(2)g)/3`
    D
    `(2m_(1)g)/3`

  • Effect of friction between pulley and thread : In ideal cases i.e., when pulley and strings are massless and no friction exists at any contact surface, then tension in the string is constant throughout its length. But consider a, massless pulley and massless string but friction exists between pulley and string With coefficient mu . Then tension at the two ends of the pulley will be different. As shown in figure, consider an element of string : dN=(T+dT) "sin" (d theta)/2+T "sin"(d theta)/2 (T+dT)"cos"(d theta)/2 -T"cos"(d theta)/2-mu dN=dr a=0 (massless strings) dT"cos"(d theta)/2=mu dN dT"cos"(d theta)/2=mu [(T+dt) "sin" (d theta)/2+T "sin" (d theta)/2] dt."cos" (d theta)/2=mu[T "sin" (d theta)/2+dT. "sin" (d theta)/2+T "sin" (d theta)/2] dT=mu[T. (d theta)/2+0+T(d theta)/2]=mu T cos theta int_(T_(1))^(T_(2)) (d T)/T=int _(0)^(pi)mu d theta rArr ln(T_(2)/T_(1))=mu pi rArrT_(2)/T_(1)=e^(mu pi) Suppose cofficient of friction between the string and pulley is mu =1/pi The tension on side of lighter mass will be:

    A
    `m_(1)g`
    B
    `m_(2)g`
    C
    `(2m_(2)g)/3`
    D
    `(4m_(1)g)/3`

  • Effect of friction between pulley and thread : In ideal cases i.e., when pulley and strings are massless and no friction exists at any contact surface, then tension in the string is constant throughout its length. But consider a, massless pulley and massless string but friction exists between pulley and string With coefficient mu . Then tension at the two ends of the pulley will be different. As shown in figure, consider an element of string : dN=(T+dT) "sin" (d theta)/2+T "sin"(d theta)/2 (T+dT)"cos"(d theta)/2 -T"cos"(d theta)/2-mu dN=dr a=0 (massless strings) dT"cos"(d theta)/2=mu dN dT"cos"(d theta)/2=mu [(T+dt) "sin" (d theta)/2+T "sin" (d theta)/2] dt."cos" (d theta)/2=mu[T "sin" (d theta)/2+dT. "sin" (d theta)/2+T "sin" (d theta)/2] dT=mu[T. (d theta)/2+0+T(d theta)/2]=mu T cos theta int_(T_(1))^(T_(2)) (d T)/T=int _(0)^(pi)mu d theta rArr ln(T_(2)/T_(1))=mu pi rArrT_(2)/T_(1)=e^(mu pi) Suppose cofficient of friction between the string and pulley is mu =1/pi If m_(2)=2em then acceleration of each mass is :

    A
    `g`
    B
    `g//3`
    C
    `eg//3`
    D
    zero

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