#linear speed

Vector Product and Rotational Kinematics

A solid sphere of mass 3kg and radius 0.1m rolls without slipping down an incline. If its linear speed is 5m/s, what is

For rolling without slipping: v = ωr. v = 5m/s, r = 0.1m. ω = vr = 50.1 = 50rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 50 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

A solid cylinder of mass 5kg and radius 0.4m rolls without slipping with a linear speed of 2.4m/s. What is its angular v

v = ωr. v = 2.4m/s, r = 0.4m. ω = vr = 2.40.4 = 6rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rigid Body Motion and Equilibrium

A particle rotates about a fixed axis at a radius of 0.3m with a linear speed of 6m/s. What is its angular velocity?

v = ωr. v = 6m/s, r = 0.3m. ω = vr = 60.3 = 20rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 20 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rigid Body Motion and Equilibrium

A thin ring of mass 2kg and radius 0.7m rolls without slipping with a linear speed of 3.5m/s. What is its angular veloci

v = ωr. v = 3.5m/s, r = 0.7m. ω = vr = 3.50.7 = 5rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rigid Body Motion and Equilibrium

A solid cylinder of mass 6kg and radius 0.5m rolls without slipping with an angular velocity of 4rad/s. What is its line

v = ωr. ω = 4rad/s, r = 0.5m. v = 4×0.5 = 2m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.0 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Kinetic Energy and Dynamics

A particle in a rigid body rotates at a distance of 0.25m from a fixed axis with an angular velocity of 12rad/s. What is

v = ωr. ω = 12rad/s, r = 0.25m. v = 12×0.25 = 3m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.0 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Kinetic Energy and Dynamics

A hollow cylinder of mass 5kg and radius 0.4m rolls without slipping with an angular velocity of 8rad/s. What is its lin

v = ωr. ω = 8rad/s, r = 0.4m. v = 8×0.4 = 3.2m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.2 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Kinetic Energy and Dynamics

A particle rotates about a fixed axis at a radius of 0.9m with a linear speed of 4.5m/s. What is its angular velocity?

v = ωr. v = 4.5m/s, r = 0.9m. ω = vr = 4.50.9 = 5rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Kinetic Energy and Dynamics

A particle in a rigid body rotates at a distance of 0.7m from a fixed axis with an angular velocity of 6rad/s. What is i

v = ωr. ω = 6rad/s, r = 0.7m. v = 6×0.7 = 4.2m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.2 m/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Kinetic Energy and Dynamics

A solid sphere of mass 2kg and radius 0.3m rolls without slipping with an angular velocity of 10rad/s. What is its linea

v = ωr. ω = 10rad/s, r = 0.3m. v = 10×0.3 = 3m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.0 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Kinetic Energy and Dynamics

A particle rotates about a fixed axis at a radius of 0.6m with a linear speed of 7.2m/s. What is its angular velocity?

v = ωr. v = 7.2m/s, r = 0.6m. ω = vr = 7.20.6 = 12rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 12 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Torque and Angular Momentum

A particle in a rigid body rotates about a fixed axis at a distance of 0.8m with an angular velocity of 5rad/s. What is

Linear speed: v = ωr. ω = 5rad/s, r = 0.8m. v = 5×0.8 = 4m/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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