Answer:
2N
Explanation:
[tex]m = m_1 = m_2 = 1 kg[/tex]
The torques generated by tangential forces on the 2 wheels are
[tex]T_1 = F_1R_1 = 1*0.5 = 0.5 Nm[/tex]
[tex]T_2 = F_2R_2 = F_2*1 = F_2 Nm[/tex]
According to Newton's 2nd law, the angular accelerations generated by these torque would be
[tex]\alpha_1 = \frac{T_1}{I_1} = \frac{0.5}{mR_1^2} = \frac{0.5}{1*0.5^2} = 2 rad/s^2[/tex]
[tex]\alpha_2 = \frac{T_2}{I_2} = \frac{F_2}{mR_2^2} = \frac{F_2}{1*1^2} = F_2 rad/s^2[/tex]
For the 2nd wheel to have the same angular acceleration, its force must be
[tex]\alpha_1 = \alpha_2[/tex]
[tex]2 = F_2[/tex]
[tex]F_2 = 2N[/tex]
The force F₂ must be 2N in order to impart identical angular accelerations on each wheel.
Given information:
for wheel 1,
mass m₁ = 1kg
radius r₁ = 0.5m
force applied F₁ = 1N
for wheel 2,
mass m₂ = 1kg
radius r₂ = 1m
force applied F₂ = to be determined
the torque on wheel 1 will be;
I₁α₁ = r₁ × F₁ .........................(1)
and the torque on wheel 2 will be:
I₂α₂ = r₂ × F₂ ...................(2)
According to the question let the angular acceleration α₁ = α₂
Dividing the eq(1) by eq(2) we get:
[tex]\frac{I_1}{I_2}= \frac{r_1\times F_1}{r_2\times F_2}\\\\ \frac{m_1r_1^2}{m_2r_2^2}= \frac{r_1\times F_1}{r_2\times F_2}\\\\ \frac{m_1r_1}{m_2r_2}= \frac{F_1}{F_2}\\\\ \frac{1\times0.5}{1\times1}= \frac{1}{F_2}[/tex]
F₂ = 2N
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