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A uniform rod AB of mass m and length l ...

A uniform rod `AB` of mass `m` and length `l` is suspended by two massless and inextensible strings `AC` and `BD` whose ends `C` and `D` are fixed as shown. Find the tension in the string `BD` immediately after the string at `A` is cut.

Text Solution

Verified by Experts

The correct Answer is:
`(15mg)/52`
`T cos 37^@=ma_(x)`
`:. a_(x)=(4T)/(5m)`…..i
`mg-Tsin37^@=ma_(y)`
`:. a_(x)=g-(3T)/(5m)`……ii
and taking torque about CM of rod,
`Tsin37^@ (l/2)=(3l^(2))/12alpha`
`:. al=(18T)/(5m)`…..iii

Now, since the sting `BD` neither slacks nor breaks, therefore the acceleration of end `B` along the a string must be zero.
`i.e. a_(y)cos37^@+(al)/2cos53^@-a_(y)cos53^@=0`


From eqn i, ii and iii
`(4T)/(5m).4/5+(18T)/(5mxx2).3/5-(g-(3T)/(5m))3/5=0`
`(52T)/(25m)(3g)/5`
`:. T(15mg)/52`
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