A disc is rolling (without slipping) on a horizontal surface C is its centre and Q and P are two points equidistant from C. Let vP, vQ and vC be the magnitude of velocities of points P, Q and C respectively, then
1. \[ V_{Q}> V_{C}> V_{P} \]
2. \[ V_{Q}< V_{C}< V_{P} \]
3. \[ V_{Q}= V_{P }, V_{C}= \frac{1}{2} V_{P} \]
4. \[ V_{Q}< V_{C}> V_{P} \]
View Answer
During pure rolling the point on ground acts as instantaneous axis of rotation. Distance from the point of contact with ground determines speed of point. so,
\[ V_{Q}> V_{C}> V_{P} \]
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is
1. Ring
2. Solid Sphere
3. Hollow Sphere
4. Disc
View Answer
\[ \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}= mgh =mg\frac{3v^{2}}{4g}= \frac{3}{4}mv^{2} \]
\[ \frac{1}{2}I\omega^{2}= \frac{1}{4}mv^{2} \]
Solving I = MR²/2 so, Object is a disc or hollow cylinder.
A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is
1. Solid cylinder
2. Solid sphere
3. Disc
4. Ring
View Answer
Given, rotational kinetic energy is 40% of total energy. so,
\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]
Solving ,
\[ I = \frac{2}{5}mR^{2} \]
Object is Solid Sphere.
Assertion (A): A disc rolls without slipping on a fixed rough horizontal surface. Then there is no point on the disc whose velocity is in vertical direction.
Reason (R): Rolling motion can be taken as combination of translation and rotation. Due to the translational part of motion a velocity (translational component) exist in horizontal direction for any point on the disc rolling on a fixed rough horizontal surface.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
The velocity of any point on a rolling disc is \( \vec{v} = v_{CM}\hat{i} + (\vec{\omega} \times \vec{r}) \). For pure rolling, \( v_{CM} = \omega R \). If \( v_x = v_{CM} + \omega y = 0 \), this occurs only at the contact point (\( y = -R \)), where \( v_y = -\omega x = 0 \). Thus, no point has a purely vertical velocity. Both A and R are true, and R explains A.
Assertion (A): By definition, pure rolling of a body occurs when velocity of its point of contact is zero relative to the surface on which it rolls.
Reason (R): A body is purely rolling (rolling without slipping). The velocity of point of contact (of body) must be zero with respect to ground.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Pure rolling is fundamentally defined by the condition that there is no relative motion (slipping) between the contact point of the rolling body and the surface it rolls on. This means their relative velocity must be zero.
Both A and R state this definition/condition, with R reinforcing A. Thus, both A and R are true, and R correctly explains A.
Assertion (A): When the body is rolling purely, the velocity of the point of contact should be zero relative to the surface in contact.
Reason (R): Friction is necessary for a body to roll purely on a level horizontal ground.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion A is the definition of pure rolling (no slipping at the contact point).
Reason R states that friction is required for a body to initiate or maintain pure rolling on a horizontal surface by providing the necessary torque/force.
Both A and R are true statements, but R describes a condition for pure rolling, not an explanation of its definition.
Assertion (A): Speed of any point on a rigid body in pure rolling can be calculated by expression \(v = r\omega\), where \(r = \text{distance of points from instantaneous centre of rotation}\).
Reason (R): Pure rolling of rigid body can be considered as a pure rotation about instantaneous centre of rotation.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
In pure rolling, the point of contact is the instantaneous center of rotation. The velocity of any point on the rigid body is given by \(v = r\omega\) where \(r\) is its distance from the ICOR. Pure rolling is essentially rotation about the ICOR.
Assertion (A): A sphere rolls down a rough inclined plane without slipping. It gains rotational K.E due to friction.
Reason (R): In this situation, work done by static friction is negative.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
For pure rolling, the point of contact is instantaneously at rest. Therefore, static friction does no work.
The static friction provides the necessary torque for angular acceleration and gain in rotational K.E.
Assertion (A): When a sphere is rolls on a horizontal table it slows down and eventually stops.
Reason (R): When the sphere rolls on the table, both the sphere and the surface deform near the contact. As a result, the normal force does not pass through the centre and provide an angular deceleration.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
A rolling sphere stops due to rolling friction. This friction arises from deformations at the contact point, causing the normal force to produce a torque that opposes the rolling motion, leading to angular deceleration.
A sphere is performing pure rolling on a rough horizontal surface with constant angular velocity.
Assertion (A): Frictional force acting on the sphere is zero.
Reason (R): Velocity of contact point is zero.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true; if a sphere rolls purely with constant angular velocity, no force (including friction) is needed to maintain its motion.
Reason (R) is true; for pure rolling, the contact point is instantaneously at rest.
However, R is a definition of pure rolling, not the explanation for zero friction in this specific case (due to zero acceleration).