Tag: physics

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Questions Related to physics

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A ball rolling off the top of a staicase of each step with height H and width W, with an initial velocity U will just hit nth step. Then n = 

  1. $\frac{2U^2H^2}{gW}$

  2. $\frac{2U^2H^2}{gW^2}$

  3. $\frac{2U^2H}{gW^2}$

  4. $\frac{2UH^2}{gW^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ball hits the nth step when its vertical displacement is nH and horizontal displacement is nW. Using y = (1/2)gt^2 and x = Ut, we get nH = (1/2)g(nW/U)^2. Solving for n gives n = 2U^2H / (gW^2).

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A small charged ball of mass m and charge q is suspended from the highers point of a ring of radius R by means of an insulated code of negligible mass.The ring is made of a rigid wire of negligible cross-section and lies in a vertical plane.On the ring, there is uniformly distributed charge Q of the same as that of q .determine the length of the cord so as the equilibrium position of the ball lies on the symmetry axis ,perpendicular to the plane of the ring. 

  1. $\left( \cfrac { 2kQqR }{ mg } \right) ^{ 1/3 }$

  2. $\left( \cfrac { kQqR }{ mg } \right) ^{ 1/3 }$

  3. $\left( \cfrac { kQqR }{ 2mg } \right) ^{ 1/3 }$

  4. $\left( \cfrac { kQqR }{ mg } \right) ^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A wheel whose radius is $r$ and moment of inertia about its-own axis is $I$, can rotate freely about its own horizontal axis. A rope is wrapped on the wheel. A boy of mass $m$ is suspended from the free end of the rope. The body is released from rest. The velocity of the body after falling a distance $h$ would be- 

  1. $\left(\dfrac{mgh}{I}\right)^{{1}/{2}}$

  2. $\left(\dfrac{2mgh}{m+I}^{{1}/{2}}\right)$

  3. $\left(\dfrac{2mgh}{m+I/r^2}\right)^{{1}/{2}}$

  4. $\left(\dfrac{m +I}{mgh}\right)^{{1}/{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of energy: mgh = (1/2)mv^2 + (1/2)I(omega)^2. Since v = omega*r, omega = v/r. So mgh = (1/2)mv^2 + (1/2)I(v/r)^2. Solving for v gives v = sqrt(2mgh / (m + I/r^2)).

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A solid cylinder (SC),Hollow cylinder (HC)& solid sphere (SS)of same mass & radii are released simultaneously from the same height on an incline. The order in which they will reach the bottom is (From least time to most time order)

  1. SC,HC,SS

  2. SS,SS,HC

  3. SS,SC,HC

  4. HC,SC,SS
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Acceleration a = g*sin(theta) / (1 + I/MR^2). The object with the smallest I/MR^2 has the largest acceleration and reaches the bottom first. I/MR^2 values: SS (2/5), SC (1/2), HC (1). Order: SS, SC, HC.

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A solid homogeneous cylinder of height h and base radius r is kept vertically on a conveyer belt moving horizontally with an increasing velocity $v=a+{ bt }^{ 2 }$. If the cylinder is not allowed to slip then the time when the cylinder is about to topple, will be equal to

  1. $\dfrac { 2rg }{ bh } $

  2. $\dfrac { rg }{ bh } $

  3. $\dfrac { 2bg }{ rh } $

  4. $\dfrac { rg }{ 2bh } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Toppling occurs when the torque due to pseudo-force (ma) about the edge exceeds the torque due to gravity. The condition is m*a*h/2 = m*g*r. With a = d^2x/dt^2 = 2bt, we get m*(2bt)*h/2 = m*g*r, so t = rg/bh.

Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A disc of radius R rolls on a horizontal surface with linear velocity $ \overrightarrow {v} = v \hat {i} $ and angular velocity $ \overrightarrow {\omega} = - \omega \hat k $ there is a particle P on the circumference of the disc which has velocity in vertical direction. the height of that particle from the ground will be

  1. $ R + \dfrac {v}{ \omega} $

  2. $ R - \dfrac {v}{ \omega} $

  3. $ R + \dfrac {R}{ 2} $

  4. $ R - \dfrac {R}{ 2} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The velocity of a point on the rim is v_p = v_cm + omega x r. For vertical velocity, the horizontal components must cancel. The height y = R - v/omega is a standard result for the point where horizontal velocity is zero.