Question

F mi of a body is 2.5 kgm² , then torque required to produce an angular acceleration of 18 rad/ s² in the body is ---

Answer 1

Given :

⟶ Moment of inertia = 2.5kgm²

⟶ Angular acc. = 18rad/s²

To Find :

⟹ Torque acts on the body

Solution :

✴ For linear motion :

$\bigstar\bf\:F=ma$

Similarly,

✴ For rotational motion :

$\bigstar\bf\:\tau=I\alpha$

where,

◕ $\tau$ denoted torque

◕ I denotes moment of inertia

◕ α denoted angular acceleration

$\dashrightarrow\tt\:\tau=I\alpha\\ \\ \dashrightarrow\tt\:\tau=2.5\times 18\\ \\ \dashrightarrow\underline{\boxed{\bf{\tau=45\:Nm}}}$

Additional Information :

• A couple produces rotation without translation.
• The MI of a rigid body depends on the mass of the body, its shape and size, distribution of mass about the axid of rotation and the position and orientation of the axis of rotation.
• Newton's second law for rotation about a fixed axis defined as the angular acceleration is directly proportional to the applied torque and is inversely proportional to the MI.

Answer 2

▪ Given -

F mi of a body is 2.5 kgm² then torque required to produce an angular acceleration of 18 rad/ s²

Here,

Moment of Inertia = 2.5 kgm2

Angular acceleration = 18 rad / s2

▪ To Find -

Torque required to produce an angular acceleration.

▪ Solution -

Here, To find Formula for torque required to produce an angular acceleration will Become,

★ T = a × i

Where,

T = Torque Required

a = Angular acceleration (Alpha)

i = Moment Of Inertia

Placing Values according to given conditions,

➯ T = a × i

➯ T = 18 × 2.5

➯ T = 45 Nm

⛬ Torque required to produce angular acceleration is 45Nm.