Q.
The position vector r→ of a particle of mass m is given by the following equation r→(t)=αt3i^+βt2j^ , where α=103ms−3 , β=5ms−2 and m=0.1 kg. At t=1 s, which of the following statement (s) is (are) true about the particle?
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a
The velocity v→ is given by v→=10i^+10j^ms−1
b
The angular momentum L→ with respect to the origin is given by L→=−5/3k^Nms
c
The force F→ is given by F→=i^+2j^N
d
The torque τ→ with respect to the origin is given by τ→=−20/3k^Nm
answer is A.
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Detailed Solution
r→ =αt3i^+βt2j^→1Velocity, v→=dr→dt=3αt2i^+2βt j^→2Acceleration, a→=dv→dt=6αti^+2βj^→3 At t=1sec From (2): v¯=3αi^+2βj^=10i^+10j^=r→×v→mL→=αt3i^+βt2j^×3αt2i^+2βt j^0.1⇒L→=2αβt4k^−3αβt4k^0.1=0.1αβt4−k^=−53t4k^→4 At t=1=−53k^The force F→=ma→From (3), F→=0.120i^+10j^ =2i^+j^=dL→dt=203t3−k^=−203k^
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