A solid sphere of radius r=2m is under a combination of rotational and translational motion as shown in figure.
If acceleration of COM (O) is ao=2m/s2 and acceleration of point A is aA=6m/s2, then find angular acceleration α.
A
4rad/s2
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B
6rad/s2
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C
2rad/s2
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D
1rad/s2
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Solution
The correct option is C2rad/s2 Given: aA=6m/s2;a0=2m/s2 As we know that, →aA=→aCOM+(→α×→rOA) ⇒→aA=→a0+(→α×→rOA) ⇒→aA−→a0=→α×→rOA....(i) Here, →aA=6^im/s2 →a0=2^im/s2 →α×→rOA=(−α^k)×(2^j) ′α′ being the magnitude of angular acceleration. ⇒→α×→rOA=−2α(^k×^j)=+2α^irad/s2 On substituting in Eq.(i) we get, 6^i−2^i=2α^i On equating the coefficients of ^i both sides: ⇒4=2α Or, α=2rad/s2 ∴ option (c) is correct.
Alternate solution: We know that acceleration of top most point in rolling motion is aA=a0+rα ∵ The tangential acceleration (rα) of point A and acceleration of centre of mass(a0), both are along the same direction (i.e +vex− direction). ⇒6=2+2α ∴α=2rad/s2