Question
Obtain an expression for torque acting on a body rotating with uniform angular acceleration.
✓
Answer
Expression for torque acting on a rotating body
a) Suppose a rigid body consists of n particles of masses $m_1, m_2, m_3,$ ......, mn which are situated at distances $r_1, r_2, r_3,$ …, rn respectively, from the axis of rotation as shown in figure
b) Each particle revolves with angular acceleration α
c) Let$ F_1, F_2, F_3,$ …., Fn be the tangential force acting on particles of masses, $m_1, m_2, m_3,$ …, mn respectively.
d) Linear acceleration of particles of masses $m_1, m_2,$…, mn are given by, $a_1 = r_1\alpha , a_2 = r_2\alpha , a_3 = r_3\alpha = rn\alpha$
e) Magnitude of force acting on particle of mass m1 is given by $F_1 = m_1a_1 = m_1r_1\alpha ( a = r\alpha )$
Magnitude of torque on particle of mass m1 is given by,
$t_1 = F_1 r_1 sin Θ$ ($\because$ Radius vector is $\perp ^{ar}$ to tangential force)
\(\therefore \quad \tau_1= F _1 r _1 \sin 90^{\circ}\)
\(= F _1 r _1\)
\(=m_1 a_1 r_1\)
\(\therefore \quad \tau_1= m _1 r _1^2 \alpha \quad\left(\because a _1= r _1 \alpha\right)\)
Similarly,
\(\tau_2=m_2 r_2^2 \alpha\)
\(\tau_3= m _3 r _3^2 \alpha\),
\(\tau_n=m_n r_n^2 \alpha\)
f. Total torque acting on the body,
\(\tau=\tau_1+\tau_2+\tau_3+\ldots .+\tau_n\)
\(\therefore \quad \tau=m_1 r_1^2 \alpha+m_2 r_2^2 \alpha+m_3 r_3^2 \alpha+\ldots+m_n r_n^2 \alpha\)
\(\therefore \quad \tau=\left(m_1 r_1{ }^2+m_2 r_2{ }^2+m_3 r_3^2+\ldots .+m_n r_n^2\right) \alpha\)
\(\therefore \quad \tau=\left(\sum_{ i =1}^{ n } m _{ i } r _{ i }^2\right) \alpha\)
But, \(\sum_{ i =1}^{ n } m _{ i } r _{ i }{ }^2= I\)
\(\therefore \quad \tau= I \alpha\)
g. If \(\alpha=1 rad / s ^2\) then \(\tau= I\).
Thus, when a torque rotates the body with uniform angular acceleration of $1 rad/s^2$ then M.I of the body about a given axis of rotation becomes equal to torque acting on it.
a) Suppose a rigid body consists of n particles of masses $m_1, m_2, m_3,$ ......, mn which are situated at distances $r_1, r_2, r_3,$ …, rn respectively, from the axis of rotation as shown in figure
b) Each particle revolves with angular acceleration α
c) Let$ F_1, F_2, F_3,$ …., Fn be the tangential force acting on particles of masses, $m_1, m_2, m_3,$ …, mn respectively.
d) Linear acceleration of particles of masses $m_1, m_2,$…, mn are given by, $a_1 = r_1\alpha , a_2 = r_2\alpha , a_3 = r_3\alpha = rn\alpha$
e) Magnitude of force acting on particle of mass m1 is given by $F_1 = m_1a_1 = m_1r_1\alpha ( a = r\alpha )$
Magnitude of torque on particle of mass m1 is given by,
$t_1 = F_1 r_1 sin Θ$ ($\because$ Radius vector is $\perp ^{ar}$ to tangential force)
\(\therefore \quad \tau_1= F _1 r _1 \sin 90^{\circ}\)
\(= F _1 r _1\)
\(=m_1 a_1 r_1\)
\(\therefore \quad \tau_1= m _1 r _1^2 \alpha \quad\left(\because a _1= r _1 \alpha\right)\)
Similarly,
\(\tau_2=m_2 r_2^2 \alpha\)
\(\tau_3= m _3 r _3^2 \alpha\),
\(\tau_n=m_n r_n^2 \alpha\)
f. Total torque acting on the body,
\(\tau=\tau_1+\tau_2+\tau_3+\ldots .+\tau_n\)
\(\therefore \quad \tau=m_1 r_1^2 \alpha+m_2 r_2^2 \alpha+m_3 r_3^2 \alpha+\ldots+m_n r_n^2 \alpha\)
\(\therefore \quad \tau=\left(m_1 r_1{ }^2+m_2 r_2{ }^2+m_3 r_3^2+\ldots .+m_n r_n^2\right) \alpha\)
\(\therefore \quad \tau=\left(\sum_{ i =1}^{ n } m _{ i } r _{ i }^2\right) \alpha\)
But, \(\sum_{ i =1}^{ n } m _{ i } r _{ i }{ }^2= I\)
\(\therefore \quad \tau= I \alpha\)
g. If \(\alpha=1 rad / s ^2\) then \(\tau= I\).
Thus, when a torque rotates the body with uniform angular acceleration of $1 rad/s^2$ then M.I of the body about a given axis of rotation becomes equal to torque acting on it.
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