As shown in the figure, wheel A of radius rA=10 cm is coupled by a belt B to the wheel C of radius rC=25 cm . The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad s- 2 . Time after which wheel C reaches a rotational speed of 100 rpm , assuming the belt does not slip, is nearly

Question

As shown in the figure, wheel A of radius rA=10 cm is coupled by a belt B to the wheel C of radius rC=25 cm . The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad s- 2 . Time after which wheel C reaches a rotational speed of 100 rpm , assuming the belt does not slip, is nearly <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-rfom3zm9li8w.png" /> (A) 4 s (B) 8 s (C) 12 s (D) 16 s

Tardigrade Admin · Accepted Answer

at=α A rA=α CrC αC=αA((rA/rC))=1.6 × (10/25)=0.64 mathrmrad / mathrms2 t=(ω C/α C)=((100 × 2 π /60)/0 .64)= text16.35 s

$4 s$

$8 s$

$12 s$

$16 s$

$a_{t} = α_{A} r_{A} = α_{C} r_{C}$
$α_{C} = α_{A} (\frac{r _{A}}{r _{C}}) = 1.6 \times \frac{10}{25} = 0.64 rad / s^{2}$
$t = \frac{ω _{C}}{α _{C}} = \frac{\frac{100 \times 2 π}{60}}{0.64} = 16.35 s$

4s

8s

12s

16s

at​=αA​rA​=αC​rC​ αC​=αA​(rC​rA​​)=1.6×2510​=0.64rad/s2 t=αC​ωC​​=0.6460100×2π​​=16.35 s

$4 s$

$8 s$

$12 s$

$16 s$

$a_{t} = α_{A} r_{A} = α_{C} r_{C}$
$α_{C} = α_{A} (\frac{r _{A}}{r _{C}}) = 1.6 \times \frac{10}{25} = 0.64 rad / s^{2}$
$t = \frac{ω _{C}}{α _{C}} = \frac{\frac{100 \times 2 π}{60}}{0.64} = 16.35 s$