A door is hinged at one end and is free to rotate about a vertical axis. Does its weight cause any torque about this axis? Give reason for your answer.

3 Views|Posted 9 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

9 months ago

This is a short answer type question as classified in NCERT Exemplar

Where weight of the door acts along negative y axis

A force can produce torque only along a direction normal to itself as $τ = r * F$ . So when the door is in the xy plane the torque produced by gravity can only along z direction. Never about

Upvote

Similar Questions for you

A circular disc reaches from top to bottom of an inclined plane of length ‘L’ when it slips down the plane, it takes time ‘t₁’. When it rolls down the plane, it takes time t₂. The value of $\frac{t_{1}}{t_{2}} i s \sqrt[]{\frac{3}{x}}$  The value of x will be__________.

Alok Kumar Singh

Case – I : When disk slides down

$t_{1} = \sqrt[]{\frac{2 L}{g s i n θ}} . . . . . . . . . . (i)$

Case – II : When disk rolls down

$t_{2} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + β^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{k}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{R / \sqrt[]{2}}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + \frac{1}{2}}}} = \sqrt[]{\frac{4 L}{3 g s i n θ}} . . . . . . (i i)$

$\Rightarrow \frac{t_{1}}{t_{2}} = \sqrt[]{\frac{2 L}{g s i n θ}} \times \sqrt[]{\frac{3 g s i n θ}{4 L}} = \sqrt[]{\frac{3}{2}}$

With reference to Fig. of a cube of edge a and mass m, state whether the following are true or false. (O is the centre of the cube.)

(a) The moment of inertia of cube about z-axis is I_z = I_x + I_y

(b) The moment of inertia of cube about z′ is I_z =I_z +ma²/2

(c) The moment of inertia of cube about z″ is I_z+ma²/2 (d) I_x = I_y

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

a, b, d

a) according to the perpendicular axes theorem statement 1 is wrong

b) As z’|z so distance between them = a $\frac{\sqrt[]{2}}{2} = \frac{a}{\sqrt[]{2}}$

So according to parallel axes theorem I_z’=I_z+m (a/ $\sqrt[]{2}$ )²= I_z+ma²/2

Hence b is true

c) z’ is not parallel to z hence Pa

Figure 7.5 shows a lamina in x-y plane. Two axes z and z′ pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z′-axis than the z-axis.)

(a) Torque τ caused by F about z axis is along ˆ -k.

(b) Torque τ′ caused by F about z′ axis is along ˆ -k.

(c) Torque τ caused by F about z axis is greater in magnitude than that about z axis.

(d) Total torque is given be τ = τ + τ′.

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

b, c

a) When r>r’

Torque about z-axis t=r $\times$ F

b) t’=r’ $\times F$ which is along negative z axis

c) t_z=Fr = magnitude of torque about z axis where r is perpendicular between F and z axis so torque along positive z axis is greater than negative

The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it ?

(a) The forces may be acting radially from a point on the axis.

(b) The forces may be acting on the axis of rotation.

(c) The forces may be acting parallel to the axis of rotation.

(d) The torque caused by some forces may be equal and opposite to that caused by other forces

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

a, b, c, d

As we know torque = r $\times$ F = rFsin $θ$

a) when forces act radially angle =0 hence torque =0

b) when forces are acting on the axis of rotation r=0 torque=0

c) when forces acting parallel to the axis of rotation angle =0 so torq

Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant, r₁ and r₂ are their respective position vectors drawn from point A which is in the plane of the parallel lines.

Choose the correct options:

(a) Angular momentum l₁ of particle 1 about A is I = mvd₁

(b) Angular momentum l2 of particle 2 about A is l₂= mvr₂

(c) Total angular momentum of the system about A is l= mv(r₁+r₂)

(d) Total angular momentum of the system about A is l = mv(d₂-d₁)⊗ represents a unit vector coming out of the page. ⊗ represents a unit vector going into the page.

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

(a), (b)As we know L= r $\times$ p where r is position vector and p is the linear momentum . the direction of L is perpendicular to both r and p by right hand rule.

For particle 1

I₁=r_{1 $\times$}mv is out of the plane of the paper and perpendicu