A wheel is made to roll without slipping, towards right, by pulling a string wrapped around a coaxial spool as shown in figure. With what velocity(in m/s) the string should be pulled so that the centre of the wheel moves with a velocity of 3 m/s?
Answer
633.3k+ views
Hint:Rolling without slipping is a combination of translation and rotation where the point of contact is instantly at rest. When an object experiences pure translational motion, all of its points move with the same speed as the center of mass, which is in the same direction and with the same speed.
Complete step by step solution:
In the given figure for Position A
$\eqalign{
& {\text{V = R}}\omega \cr
& \Rightarrow \omega {\text{ = }}\dfrac{{\text{V}}}{{\text{R}}} \cr
& \Rightarrow \omega {\text{ = }}\dfrac{3}{{0.3}} \cr
& \therefore \omega {\text{ = 10}} \cr
& {\text{So, velocity of string = 3 - r}}\omega \cr
& \Rightarrow {\text{velocity of string = 3 - 1}} \cr
& \therefore {\text{velocity of string = 2}} \cr} $
Additional Information:When we slide an object across a surface (say, a book on a table), it will normally slow down rapidly due to frictional forces. When we do the same with a round object, like a water bottle, it may initially slide a bit (especially if we press it hard), but it will start to spin quickly. We can easily check that when rotating, the object loses much less kinetic energy to work than when sliding. Now we will take the same bottle of water, either on its bottom (only sliding) or sideways (a little more sliding), push it with the same initial force, and release, the rolling bottle goes much further. However, ironically, the bottle can only roll because of friction. To start rolling, we need to change its angular momentum, which requires torque, which is provided by the friction force acting on the bottle.
When a bottle (or ball, or any round object) rolls, the instantaneous velocity of the point that it touches the surface on which it rolls is zero.
Consequently, its speed of rotation $\omega$, the speed of translation of its center of rotation ${V_r}$ (where r indicates rolling) are related by
${V_r}$ = $\omega$ R
where, R is the relevant radius of our object. If the object's center of rotation is moving faster than Vr , the rotation cannot "keep up" and the object is sliding on the surface. We call this type of movement sliding. Due to friction, objects that undergo sliding motion generally slow down rapidly to Vr, at which point they roll without slipping.
Note:The wheel rolls without slipping only if there is no horizontal movement of the wheel at the contact point(relative to the surface / ground). Therefore, the contact point must also have zero horizontal movement (with respect to the surface / ground).
Complete step by step solution:
In the given figure for Position A
$\eqalign{
& {\text{V = R}}\omega \cr
& \Rightarrow \omega {\text{ = }}\dfrac{{\text{V}}}{{\text{R}}} \cr
& \Rightarrow \omega {\text{ = }}\dfrac{3}{{0.3}} \cr
& \therefore \omega {\text{ = 10}} \cr
& {\text{So, velocity of string = 3 - r}}\omega \cr
& \Rightarrow {\text{velocity of string = 3 - 1}} \cr
& \therefore {\text{velocity of string = 2}} \cr} $
Additional Information:When we slide an object across a surface (say, a book on a table), it will normally slow down rapidly due to frictional forces. When we do the same with a round object, like a water bottle, it may initially slide a bit (especially if we press it hard), but it will start to spin quickly. We can easily check that when rotating, the object loses much less kinetic energy to work than when sliding. Now we will take the same bottle of water, either on its bottom (only sliding) or sideways (a little more sliding), push it with the same initial force, and release, the rolling bottle goes much further. However, ironically, the bottle can only roll because of friction. To start rolling, we need to change its angular momentum, which requires torque, which is provided by the friction force acting on the bottle.
When a bottle (or ball, or any round object) rolls, the instantaneous velocity of the point that it touches the surface on which it rolls is zero.
Consequently, its speed of rotation $\omega$, the speed of translation of its center of rotation ${V_r}$ (where r indicates rolling) are related by
${V_r}$ = $\omega$ R
where, R is the relevant radius of our object. If the object's center of rotation is moving faster than Vr , the rotation cannot "keep up" and the object is sliding on the surface. We call this type of movement sliding. Due to friction, objects that undergo sliding motion generally slow down rapidly to Vr, at which point they roll without slipping.
Note:The wheel rolls without slipping only if there is no horizontal movement of the wheel at the contact point(relative to the surface / ground). Therefore, the contact point must also have zero horizontal movement (with respect to the surface / ground).
Recently Updated Pages
10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Difference between physical and chemical change class 11 chemistry CBSE

What are derived physical quantities Give any two examples class 11 physics CBSE

Which of the following is a derived physical quantity class 11 physics CBSE

Describe various types of epithelial tissues with the class 11 biology CBSE

