A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant power p and increases its speed. The angular acceleration of particle at time (t) is proportional

Question

A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant power p and increases its speed. The angular acceleration of particle at time (t) is proportional (A) (1/√t) (B) √t (C) t° (D) t3 / 2

Tardigrade Admin · Accepted Answer

Work =P t Using W=Δ K P t=(1/2) m(r w)2 P t=(1/2) m r2 w2 P t=(1/2) m r2 α2 t2 (∵ α=(w/t)) So α2 propto (1/t) ⇒ α propto (1/√t)

Q. A particle of mass $m$ moves in a circular path of radius $r$ , under the action of force which delivers it constant power $p$ and increases its speed. The angular acceleration of particle at time $(t)$ is proportional

$\frac{1}{t}$

$t$

$t^{\circ}$

$t^{3/2}$

Work $= Pt$
Using $W = Δ K$
$Pt = \frac{1}{2} m (r w)^{2}$
$Pt = \frac{1}{2} m r^{2} w^{2}$
$Pt = \frac{1}{2} m r^{2} α^{2} t^{2}$
$(∵ α = \frac{w}{t})$
So $α^{2} \propto \frac{1}{t}$
$\Rightarrow α \propto \frac{1}{t}$

Q. A particle of mass m moves in a circular path of radius r, under the action of force which delivers it constant power p and increases its speed. The angular acceleration of particle at time (t) is proportional

t​1​

t​

t∘

t3/2

Work =Pt Using W=ΔK Pt=21​m(rw)2 Pt=21​mr2w2 Pt=21​mr2α2t2 (∵α=tw​) So α2∝t1​ ⇒α∝t​1​

Q. A particle of mass $m$ moves in a circular path of radius $r$ , under the action of force which delivers it constant power $p$ and increases its speed. The angular acceleration of particle at time $(t)$ is proportional

$\frac{1}{t}$

$t$

$t^{\circ}$

$t^{3/2}$

Work $= Pt$
Using $W = Δ K$
$Pt = \frac{1}{2} m (r w)^{2}$
$Pt = \frac{1}{2} m r^{2} w^{2}$
$Pt = \frac{1}{2} m r^{2} α^{2} t^{2}$
$(∵ α = \frac{w}{t})$
So $α^{2} \propto \frac{1}{t}$
$\Rightarrow α \propto \frac{1}{t}$