The position vector `vec(r)` of a particle of mass m is given by the following equation `vec(r) (t) = at^(3) hat(i) + beta t ^(2) hat(j) " where " alpha = 10//3 ms^(-3),beta = 5 ms^(-2) and m = 0 . 1 kg ` . At t = 1 s, which of the following statement (s) is (are) true about the particle ?

Options

A. The velocity `vecv` is given by `vec v=(10 hati+10 hatj) ms^(-1)`
B. The angular momentum `vecL` with respect to the origin is given by `vecL=-(5//3) hatj Nms`
C. the forece `vecF`n is given by `vecF=(hati+2hatj)N`
D. The torque `vectau` with respect to the origin is given by `vectau=-(20//3)hatkNm`

Answer

A, B, D

Explanation

First let us calculate velocity of the particle at t = 1s.
`vec r= alpha t^(3) hati+ beta t^(2) hatj`
`vec v=(d vec r)/(dt)=(3 alpha t^(2)) hati+(2 beta t) hatj`
We have `t=1 sec, alpha=180//3 ms^(-3), beta= 5 ms^(-2)`
`rArr vec =10 hati+10 hatj`
ence option (a) is correct. >br> Let us now calculate angular momentum of the particle.
`vecL= m(vec r xx vec v))`
We have `vec r= alpht^(3) hati+ betat^(2) hatj and vec v=(3alpha t^(2)) hati+(2 beta t) hatj`
`rArr vecL=m[ 2alpha beta t^(4)(hatk)+3 alpha beta t^(4)(-hatk)]`
`vecL=m(alpha beta^(4))(-hatk)`
We have `t=1 sec , alpha 10//3 ms^(-3), beta =5 ms^(-2)`
`rArr vecL=(5)/(3)(-hatk)N ms`
Hence option (b) is also correct. By differentiating velocity vector we can find acceleration to know about force acting on the particle.
`vecv= (3alpha t^(2)) hati+(2 beta t)hatj`
`rArr veca=(d vecv)/(dt)=(6 alpha t)hati+(2beta) hatj`
We have `t=1 sec, alpha=10//3 ms^(-3), beta= 5 ms^(-2)
`rArr vec a=20 hati+10 hatj` Hence option (c) is wrong.
Torque can be defined as rate of change of angular momentum
`vecL= m(alpha betat^(4))(-hatk)`
`vec tau=(dvecL)/(dt)`
`=(d)/(dt)(m alpha beta t^(4))(-hatk)`
`tau=(4 m alpha beta t^(3))(-hatk)`
We have `t=1 sec, alpha, =10//3 ms^(-3) , beta m s^(-2) , m=0.1 kg`.
`tau=(20)/(3)(-hatk)N-m`
Hence option (d) is also correct.