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This is one of our many quizzes on Tangential velocity. Click on this to access more such competitive quizzes on Kinematics.

Tangential velocity Quiz 2

physics

 What will be the tangential velocity of the earth if the earth’s angular momentum is $245\,\text{kg}{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}}$ and its moment of inertia is $49\,\text{kg}{{\text{m}}^{\text{2}}}$ ? Assume that the radius of the earth is $14\times {{10}^{23}}\,\text{m}$. 

 First, calculate the value of angular velocity with the help of angular momentum and then find the value of tangential velocity. 

  $\text{Angular momentum is given by,} \\ \Rightarrow L=I\omega  \\ \text{L}=245\,\text{kg}{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}},\text{ } \\ \text{I}=49\,\text{kg}{{\text{m}}^{\text{2}}} \\ \Rightarrow \omega =\dfrac{L}{I} \\ \Rightarrow \omega =\dfrac{245}{49} \\ \Rightarrow \omega =5\,\text{rad}{{\text{s}}^{\text{-1}}} \\ \text{Now, v}=\text{r}\omega  \\ \Rightarrow v=14\times {{10}^{23}}\times 5 \\ \therefore v=70\times {{10}^{23}}\,\text{m}{{\text{s}}^{-1}}$

Suppose if a particle is moving around an object, the path obtained by that particle during revolving is like a sine-wave and the maximum distance from the object of that particle is $13\,\text{m}$ and the minimum distance is $6\,\text{m}$ . If the angular momentum for the maximum distance of the particle is $39\,\text{kg}{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}}$ then what will be the linear momentum of a moving particle along a circle. Assume that the mass of the particle is $4\text{ kg}$.

 First, calculate the value of the uniform speed of the particle around a circular path and then calculate linear momentum. 

  $\text{Angular momentum is given by,} \\ L=mvr \\ \text{According to the question:} \\ L=39\,\text{kg}{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{-1}}},\text{ }m=4\,\text{kg},\text{ }r=13\,\text{m} \\ \therefore v=\dfrac{L}{mr} \\ \Rightarrow v= \dfrac{39}{4\times 13} \\ \Rightarrow v=\dfrac{3}{4}\text{ m}{{\text{s}}^{-1}} \\ \text{Also, } \\ \text{Linear momentum(p)}=\text{mv} \\ \Rightarrow p=4\times \dfrac{3}{4} \\ \therefore p=3\,\text{kgm}{{\text{s}}^{\text{-1}}}$

 George puts his ball on a slide and then he leaves his ball and now the ball is rolling down on an inclined slide. If the rotational energy of the ball when it reaches the ground is $16\,\text{J}$ and linear velocity is $\text{4 m}{{\text{s}}^{-1}}$ then calculate the value of the height at which George put the ball on a slide. Assume that the mass of the ball is $3\,\text{kg}$. Take $g=10\,\text{m}{{\text{s}}^{\text{-2}}}$ 

 Apply the law of conservation of energy to calculate the value of height. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ce6f6d23-9908-47d9-8160-58f8a969672f1803634787183378692.png"/><br>$\text{Mass(m)}=3\,\text{kg,} \\ \text{linear velocity(v)}=4\,\text{m}{{\text{s}}^{-1}} \\ \text{Rotational Energy(R}\text{.E)}=16 \\ \text{Apply the concept of conservation of energy:} \\ \text{Kinetic Energy}+\text{Rotational Energy}=\text{Potential Energy} \\ \Rightarrow \dfrac{1}{2}M{{v}^{2}}+R.E.=Mgh \\ \Rightarrow \dfrac{1}{2}\times 3\times {{4}^{2}}+16=3\times 10\times h \\ \Rightarrow 40=30h \\ \therefore h=1.33\,\text{m}$

If a circular object moving with the horizontal speed for a function of $x={{t}^{3}}+2t$, where $x$ is referred to the distance and $t$ refers to the time, then calculate the value of the angular velocity at the time $4\text{ sec}$, if the radius of the circular object is $5\,\text{m}$. 

 First, calculate the value of linear velocity and then calculate angular velocity. 

 $\text{Given function for the circular object is,} \\ x={{t}^{3}}+2t \\ \Rightarrow v=\dfrac{dx}{dt} \\ \Rightarrow v= \dfrac{d}{dt}\left( {{t}^{3}}+2t \right) \\ \Rightarrow v= 3{{t}^{2}}+2 \\ \text{Also,} \\ \text{time(t) }=\text{ 4 sec} \\ \Rightarrow v=3\times {{\left( 4 \right)}^{2}}+2=50 \\ \text{Radius(r) given}=\text{5}\,\text{m} \\ \text{Now v}=r\omega  \\ \omega =\dfrac{v}{r} \\ \Rightarrow \omega = \dfrac{50}{5} \\ \therefore \omega =10\,\text{rad}{{\text{s}}^{\text{-1}}}$

 Suppose a body that is attached to a string and that string is wound around a circular cylinder.After that, if the mass of the attached body(m) is reduced by half and the radius of the cylinder is increased by two times then what will be the new tangential velocity of the body? Assume that the cylinder starts rotating about its axis of rotation and the mass of the cylinder is M. 

 First, calculate the moment of inertia of the cylinder and then calculate the linear acceleration of the body. After that calculate the tangential velocity of the body. 

 $\text{Moment of Inertia of cylinder,} \\ I=\dfrac{1}{2}M{{R}^{2}} \\ \text{here,}\,R=2R \\ \Rightarrow I=\dfrac{1}{2}M4{{R}^{2}} \\ \Rightarrow I=2M{{R}^{2}} \\ \text{Also,} \\ \text{Linear acceleration of a point mass(a):} \\ \Rightarrow a=\dfrac{g}{1+\dfrac{I}{m{{R}^{2}}}} \\ \text{here},\,m\left( \text{mass of the body attached to a string} \right)=\dfrac{m}{2} \\ \therefore a=\dfrac{g}{1+\dfrac{2M{{R}^{2}}}{\dfrac{m}{2}4{{R}^{2}}}} \\ \Rightarrow a= \dfrac{g}{1+\dfrac{M}{m}} \\ \text{Also,} \\ v=u+at \\ \Rightarrow v=0+\dfrac{gt}{1+\dfrac{M}{m}} \\ \therefore v=\dfrac{gt}{1+\dfrac{M}{m}}$ 