A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque (in Nm ) that will make the system rotate about AB at 25 rotations per second in 5 s, is close to:

Question

A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque (in Nm ) that will make the system rotate about AB at 25 rotations per second in 5 s, is close to: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Angular acceleration, α=(ω-ω0/t)=(25 × 2 π-0/5)=10 π rad / s 2 τ =I α ⇒ τ =((5/4) m R2) α ≈((5/4))(5 × 10-3)(10-4) 10 π =2.0 × 10-5 Nm

Q. A metal coin of mass $5 g$ and radius $1 c m$ is fixed to a thin stick $A B$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque (in $N m$ ) that will make the system rotate about $A B$ at $25$ rotations per second in $5 s$ , is close to:

Angular acceleration,
$α = \frac{ω - ω _{0}}{t} = \frac{25 \times 2 π - 0}{5} = 10 π r a d / s^{2}$
$τ = I α$
$\Rightarrow τ = (\frac{5}{4} m R^{2}) α \approx (\frac{5}{4}) (5 \times 1 0^{- 3}) (1 0^{- 4}) 10 π$
$= 2.0 \times 1 0^{- 5} N m$

Q. A metal coin of mass 5g and radius 1cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque (in Nm ) that will make the system rotate about AB at 25 rotations per second in 5s, is close to:

Angular acceleration, α=tω−ω0​​=525×2π−0​=10πrad/s2 τ=Iα ⇒τ=(45​mR2)α≈(45​)(5×10−3)(10−4)10π =2.0×10−5Nm

Q. A metal coin of mass $5 g$ and radius $1 c m$ is fixed to a thin stick $A B$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque (in $N m$ ) that will make the system rotate about $A B$ at $25$ rotations per second in $5 s$ , is close to:

Angular acceleration,
$α = \frac{ω - ω _{0}}{t} = \frac{25 \times 2 π - 0}{5} = 10 π r a d / s^{2}$
$τ = I α$
$\Rightarrow τ = (\frac{5}{4} m R^{2}) α \approx (\frac{5}{4}) (5 \times 1 0^{- 3}) (1 0^{- 4}) 10 π$
$= 2.0 \times 1 0^{- 5} N m$