Explanation
Centripetal acceleration and tangential acceleration is normal to each other.
Centripetal acceleration, ac=v2r=20250=8ms−2
Tangential acceleration, at=0
Hence, magnitude of acceleration is 8ms−1
The acceleration will be directed towards the centre of the circular loop.
angular velocity = 2πf = 2 x 22/7 x 14/25 = 88/25 rad/s
Centripetal acceleration = rω2 = 0.8 m x 88 x 88 / 225 = 27.53 m/s2
LetAngle of projection for A=30oAngle of projection for B=60oRequired condition:Distance traveled by A−Distance traveled by B=x[u3√]cos30o×t−ucos60o×t=x[u3√](3√2)×t−u(12)×t=xut[32−12]=xut=xt=xu
Moment of inertia of uniform rod of length(l) -
I=Ml212
- wherein
About axis passing through its center & perpendicular to its length.
Law of conservation of angular moment -
→r=d→Ldt
If net torque is zero
i.e d→Ldt=0,
→L= constant
angular momentum is conserved only when external torque is zero .
Centre of mass from system from 0
=>8m×0+m(L3)−2m(L6)8m+m+2m=0
So,centre of mass is at zero.
From conservation of ang.momentum; Li=Lf
Li=m(2v)×(L3)+2mv×(L6)=mvL
Lf=[(8m)×L212+m(L3)2+2m(L6)2]ω
=[23mL2+mL29+mL218]ω
=(12+2+118)mL2ω=56mL2ω
56mL2ω=mvl,
∴ω=6v5L
A particle is moving in xy−plane in a circular path with center at the origin. If at an instant the position of the particle is given by 1√2(ˆi+ˆj), then the velocity of the particle is along
Correct answer: Option A
Hint: Differentiation of displacement gives velocity and differentiation of velocity gives acceleration with respect to time.
Explanation of correct option:
Given: Displacement - Time relation
x=(t+5)−1...................(i)
Step 1:Differentiating both side w.r.t. ′t′ of equation (i)
We get,
dxdt=ddt(t+5)−1
∴dxdt=(−1)(t+5)−2(1+0)
dxdt is velocity.
⇒v=−(t+5)−2...................(ii)
Step 2:Differentiating both side w.r.t. ′t′ of equation (ii)
dvdt=ddt[−(t+5)−2]
∴dvdt = −(−2)(t+5)−3.(1)
dvdt is acceleration.
Formatting equation (iii) to out the value of equation (ii)
⇒a=2×{−(t+5)−2}{−(t+5)−2×12}
⇒a=2v×v12
∴a=2v32
As 2 is constant so a varies with v such as
a∝v32 i.e. (acceleration)∝(velocity)32
Hence, Acceleration of a particle is proportional to (velocity)32
When a move in circular motion, centripetal force will act on the body toward center.
When body moves with uniform angular velocity in circular path.
Then,
(i) Centripetal force, {{F}_{c}}=m\omega {{l}^{2}}
(ii) Speed of the particle, v=\omega l
(iii) Because it is horizontal circle and there is no gravity. Therefore there is no force in vertical direction.
{\textbf{Step1: Identify the values that are given.}}
{\text{Radius = R}}
{\text{Static friction = }}\mu
{\text{Now, the centripetal force is providing by frictional force between the road and tyre}}
{\textbf{Step2: Find v. }}
\text{The centripetal force is given by } \dfrac{{m{v^2}}}{R}
\text{and the frictional force is } = \mu mg
\text{For avoiding car to skid the cetripetal force must be equal to frictional force }
\dfrac{{m{v^2}}}{R} = \mu mg
v = \sqrt {\mu Rg}
{\text{Hence,this is the required solution}}
{\textbf{Hence,}} {\textbf{the final answer is }} \mathbf {v =} \mathbf { { \sqrt {\mu Rg} .}}
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