Question

plzz answer it in details...a 100 point question and dont copy it from net

Answer 1

Hey there !!

r(t)= αt³i+βt²j

Differentiate r wrt to t
dr/dt=3αt²i+2βtj=v--------------------------------EQUATION 1

Given that α=10/3  β=5 t=1s
v=3(10/3)i+2*5j
v=10i+10j

Now again differentiating equaiton 1 wrt to t

dv/dt=6αti+2βj =a---------------------------------------EQUATION 2

F=ma
  
  =0.1(6(10/3)i+2(5)j)
 = 0.1(20i+10j)
Force= (2i+j)N

Now,
        Torque=r x F=m(αt³i+βt²j)x(6αti+2βj)
                            =0.1(2αβt³k-6αβt³k) 
                           =0.1(2*10*5/3i-6*10*5j/3)
                          =-20/3k Nm

Answer 2

Here, given
position vector (r ) = at³i + ßt² j
in components

x = at³
y = ßt²
where, a = 10/3 m/s³ and ß = 5m/s²
we know,
velocity is the rate of change of displacement .
Vx = dx/dt = d( at³)/dt = 3at²
at t = 1 sec, Vx = 3 × 10/3 × (1)² = 10m/s

again,
Vy = dy/dt = d(ßt²)/dt = 2ßt
at t = 1 sec , Vy = 2 × 5 × 1 = 10 m/sec

hence,
velcity in vector form,
V = 10i + 10j

so, option (A) is correct


angular momentum = m( r × V)
where r is the position vector and v is the velocity vector .
now ,
here,
r = at³i + ßt²j = 10/3i + 5j
v = 10i + 10j
m = 0.1 kg

L = 0.1 { ( 10/3i + 5j)× ( 10i + 10j )}
= 0.1 × 10 { 10/3 (i×i) + 10/3(i×j) + 5(j×i)+5(j×j)}
= { 0 + 10/3k -5k+ 0 }
= -5/3k
so, option (B) is incorrect .

Force = ma

ax = dVx/dt = d²x/dt² = 6at = 6 × 10/3 = 20
ay = dVy/dt =d²y/dt² = 2ß = 10

hence, acceleration vector ,
a = 20i + 10j

F = ma
= 0.1( 20i + 10j )
= 2i + j
hence, option (C) is incorrect .

Torque = r×F
=( 10/3i + 5j) × ( 2i + j)
= ( 10/3 k - 10k )
= -20/3 K
so, option (D) is correct .


finally option (A) and (D) are correct .