Answer
Verified
430.5k+ views
Hint: The normal force of the body will be the sum of forces acting in a downward direction. But to find the normal force we need to know the velocity of the sphere in the cup, due to its rolling motion. We can find it using the law of the conservation of energy for the sphere.
Formula used:
$N = mg + \dfrac{{m{v^2}}}{R}$
${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$
Complete answer:
In the question, they’ve asked us for the normal force acting on the sphere. Normal force ‘$N$’, being a contact force balances the weight and the driving force that keeps the sphere moving in the cup. They are mathematically given by,
$N = mg + \dfrac{{m{v^2}}}{R}$
Here,
$m$ is the mass of the sphere
$v$ is the velocity of the sphere
$R$ is the radius of the hemispherical cup
The velocity of the sphere is due to the rolling motion. It can be found as follows.
Let’s draw a rough sketch, for a better understanding and easier explanation.
Initially, the sphere is on the rim, at a height of R from the ground. So, it will have a potential energy of
$P.E. = mgR$
Where,
$P.E.$ is the potential energy of the sphere
$m$ is the mass of the sphere
$g$ is the acceleration due to gravity
$R$ is the height of the sphere from the ground
Once the sphere reaches the bottom of the cup, it has rotational kinetic energy about its center of mass and translational kinetic energy due to rolling motion. So, the total kinetic energy of the sphere can be written as
${\left( {K.E.} \right)_T} = {\left( {K.E.} \right)_{rot}} + {\left( {K.E.} \right)_{trans}}$
The rotational kinetic energy of the sphere is given by
${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$
Where,
$I$ is the moment of inertia of the sphere
$\omega $ is the angular velocity of the sphere
As the sphere is rotating about its tangential axis, we have $I = \dfrac{2}{5}m{r^2}$ and the angular velocity can be written as $\omega = \dfrac{v}{r}$. Here, the $r$ indicates the radius of the sphere. Using these relations, rotational kinetic energy can be given by
$\eqalign{
& {\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2} \cr
& \Rightarrow {\left( {K.E.} \right)_{rot}} = \dfrac{1}{2} \times \dfrac{2}{5}m{r^2} \times {\left( {\dfrac{v}{r}} \right)^2} \cr
& \Rightarrow {\left( {K.E.} \right)_{rot}} = \dfrac{1}{5}m{v^2} \cr} $
Similarly, the translational kinetic energy is given by ${\left( {K.E.} \right)_{trans}} = \dfrac{1}{2}m{v^2}$.
Substituting these values, we have the total kinetic energy of the sphere will
$\eqalign{
& {\left( {K.E.} \right)_T} = {\left( {K.E.} \right)_{rot}} + {\left( {K.E.} \right)_{trans}} \cr
& \Rightarrow {\left( {K.E.} \right)_T} = \dfrac{1}{5}m{v^2} + \dfrac{1}{2}m{v^2} \cr
& \Rightarrow {\left( {K.E.} \right)_T} = \dfrac{7}{{10}}m{v^2} \cr} $
From the law of conservation of energy, the potential energy changes into kinetic energy. Thus,
$\eqalign{
& P.E. = {\left( {K.E.} \right)_T} \cr
& \Rightarrow mgR = \dfrac{7}{{10}}m{v^2} \cr
& \Rightarrow {v^2} = \dfrac{{10}}{7}gR \cr} $
Now, the normal force of the sphere can be found by substituting this value, as
$\eqalign{
& N = mg + \dfrac{{m{v^2}}}{R} \cr
& \Rightarrow N = mg + \dfrac{{m \times \dfrac{{10}}{7}gR}}{R} \cr
& \Rightarrow N = mg + \dfrac{{10}}{7}mg \cr
& \therefore N = \dfrac{{17}}{7}mg \cr} $
Therefore, the correct option is A.
Note:
One can misjudge that normal force is always equal to the weight. The Normal force is a supporting force one the object against a stable object. In our problem, the ball or sphere is moving in a circular motion with some velocity. So, the normal force will not just counteract weight but also force due to circular motion, towards the center.
Formula used:
$N = mg + \dfrac{{m{v^2}}}{R}$
${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$
Complete answer:
In the question, they’ve asked us for the normal force acting on the sphere. Normal force ‘$N$’, being a contact force balances the weight and the driving force that keeps the sphere moving in the cup. They are mathematically given by,
$N = mg + \dfrac{{m{v^2}}}{R}$
Here,
$m$ is the mass of the sphere
$v$ is the velocity of the sphere
$R$ is the radius of the hemispherical cup
The velocity of the sphere is due to the rolling motion. It can be found as follows.
Let’s draw a rough sketch, for a better understanding and easier explanation.
Initially, the sphere is on the rim, at a height of R from the ground. So, it will have a potential energy of
$P.E. = mgR$
Where,
$P.E.$ is the potential energy of the sphere
$m$ is the mass of the sphere
$g$ is the acceleration due to gravity
$R$ is the height of the sphere from the ground
Once the sphere reaches the bottom of the cup, it has rotational kinetic energy about its center of mass and translational kinetic energy due to rolling motion. So, the total kinetic energy of the sphere can be written as
${\left( {K.E.} \right)_T} = {\left( {K.E.} \right)_{rot}} + {\left( {K.E.} \right)_{trans}}$
The rotational kinetic energy of the sphere is given by
${\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$
Where,
$I$ is the moment of inertia of the sphere
$\omega $ is the angular velocity of the sphere
As the sphere is rotating about its tangential axis, we have $I = \dfrac{2}{5}m{r^2}$ and the angular velocity can be written as $\omega = \dfrac{v}{r}$. Here, the $r$ indicates the radius of the sphere. Using these relations, rotational kinetic energy can be given by
$\eqalign{
& {\left( {K.E.} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2} \cr
& \Rightarrow {\left( {K.E.} \right)_{rot}} = \dfrac{1}{2} \times \dfrac{2}{5}m{r^2} \times {\left( {\dfrac{v}{r}} \right)^2} \cr
& \Rightarrow {\left( {K.E.} \right)_{rot}} = \dfrac{1}{5}m{v^2} \cr} $
Similarly, the translational kinetic energy is given by ${\left( {K.E.} \right)_{trans}} = \dfrac{1}{2}m{v^2}$.
Substituting these values, we have the total kinetic energy of the sphere will
$\eqalign{
& {\left( {K.E.} \right)_T} = {\left( {K.E.} \right)_{rot}} + {\left( {K.E.} \right)_{trans}} \cr
& \Rightarrow {\left( {K.E.} \right)_T} = \dfrac{1}{5}m{v^2} + \dfrac{1}{2}m{v^2} \cr
& \Rightarrow {\left( {K.E.} \right)_T} = \dfrac{7}{{10}}m{v^2} \cr} $
From the law of conservation of energy, the potential energy changes into kinetic energy. Thus,
$\eqalign{
& P.E. = {\left( {K.E.} \right)_T} \cr
& \Rightarrow mgR = \dfrac{7}{{10}}m{v^2} \cr
& \Rightarrow {v^2} = \dfrac{{10}}{7}gR \cr} $
Now, the normal force of the sphere can be found by substituting this value, as
$\eqalign{
& N = mg + \dfrac{{m{v^2}}}{R} \cr
& \Rightarrow N = mg + \dfrac{{m \times \dfrac{{10}}{7}gR}}{R} \cr
& \Rightarrow N = mg + \dfrac{{10}}{7}mg \cr
& \therefore N = \dfrac{{17}}{7}mg \cr} $
Therefore, the correct option is A.
Note:
One can misjudge that normal force is always equal to the weight. The Normal force is a supporting force one the object against a stable object. In our problem, the ball or sphere is moving in a circular motion with some velocity. So, the normal force will not just counteract weight but also force due to circular motion, towards the center.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Advantages and disadvantages of science
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference Between Plant Cell and Animal Cell
10 examples of evaporation in daily life with explanations
Give 10 examples for herbs , shrubs , climbers , creepers
Write a letter to the principal requesting him to grant class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE