Learn Unit 4-Motion of System of Particles with Free Lessons & Tips

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your question about the center of mass of various geometric shapes.

(i) For a sphere: The center of mass of a sphere lies at its geometric center, regardless of its size or density distribution. This is true for a sphere of uniform mass density.

(ii) For a cylinder: The center of mass of a uniform cylinder lies at the midpoint of its central axis, assuming the cylinder is oriented such that its axis is vertical.

(iii) For a ring: The center of mass of a ring, also known as a circular hoop, lies at its geometrical center, which coincides with the center of the circle.

(iv) For a cube: The center of mass of a uniform cube lies at the geometric center of the cube, where the diagonals intersect.

Regarding your second question, whether the center of mass necessarily lies inside the body, the answer is yes, for a body with uniform mass density. The center of mass is a weighted average of the positions of all the mass elements in the body. Since every part of the body contributes to the calculation of the center of mass, it must lie somewhere within the body itself. However, for irregularly shaped bodies or bodies with non-uniform mass distribution, the center of mass may lie outside the body.

As a seasoned tutor registered on UrbanPro, I'd be glad to assist you with this question. UrbanPro is indeed a fantastic platform for online coaching and tuition needs.

To tackle this problem, let's first understand the concept of the center of mass (CM) of a molecule. The center of mass of a system is the point where you can consider the entire mass of the system to be concentrated for the purpose of analyzing its motion. In a diatomic molecule like HCl, the center of mass lies along the line joining the two nuclei, and it divides this line in a ratio inversely proportional to the masses of the atoms.

Given that the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^-10m), and chlorine is approximately 35.5 times as massive as hydrogen, we can use the formula for the center of mass:

xCM=m1⋅x1+m2⋅x2m1+m2xCM=m1+m2m1⋅x1+m2⋅x2

Where:

• xCMxCM is the position of the center of mass,
• m1m1 and m2m2 are the masses of the two atoms (in this case, hydrogen and chlorine),
• x1x1 and x2x2 are the positions of the two atoms (nuclei).

Let's denote:

• m1m1 as the mass of hydrogen (let's denote it as mHmH),
• m2m2 as the mass of chlorine (let's denote it as mClmCl),
• x1x1 as the position of hydrogen,
• x2x2 as the position of chlorine.

Since nearly all the mass of an atom is concentrated in its nucleus, we can directly equate the masses of the atoms to their respective nuclei.

Now, substituting the given values:

mH=mmH=m mCl=35.5mmCl=35.5m x1=0x1=0 x2=1.27×10−10x2=1.27×10−10

xCM=m⋅0+(35.5m)⋅(1.27×10−10)m+35.5mxCM=m+35.5mm⋅0+(35.5m)⋅(1.27×10−10)

xCM=35.5×1.27×10−1036.5xCM=36.535.5×1.27×10−10

xCM≈4.4985×10−1036.5xCM≈36.54.4985×10−10

xCM≈1.23×10−11xCM≈1.23×10−11

So, the approximate location of the center of mass of the HCl molecule is 1.23×10−111.23×10−11 meters away from the hydrogen nucleus, along the line joining the two nuclei.

Certainly! This is a classic problem in physics involving the conservation of momentum. Let me break it down for you.

Initially, when the child is sitting stationary at one end of the trolley, the center of mass (CM) of the system (trolley + child) is simply the center of the trolley, and it moves with a speed V.

When the child gets up and starts running about on the trolley, they exert a force on the trolley in the opposite direction to their motion. According to Newton's third law, the trolley exerts an equal and opposite force on the child.

Now, the total momentum of the system remains constant since there are no external forces acting on it. However, the distribution of mass within the system changes as the child moves.

Let M be the mass of the trolley and m be the mass of the child. Initially, the total momentum is M×VM×V.

When the child starts moving, to maintain the total momentum constant, the velocity of the trolley must decrease and the velocity of the child must increase.

Let VtVt be the final velocity of the trolley and VcVc be the final velocity of the child relative to the trolley.

The final momentum of the system is (M×Vt)+(m×Vc)(M×Vt)+(m×Vc).

Since momentum is conserved, we have:

M×V=(M×Vt)+(m×Vc)M×V=(M×Vt)+(m×Vc)

Now, to find the velocity of the center of mass of the system, we need to consider that the center of mass of the system moves with the same velocity as if all its mass were concentrated at that point.

The total mass of the system is M + m, and the velocity of the center of mass (V_cm) is given by:

Vcm=(M×Vt)+(m×Vc)M+mVcm=M+m(M×Vt)+(m×Vc)

Thus, to find the velocity of the center of mass of the system, we need to solve for VcmVcm using the equation derived above.

Certainly! UrbanPro is indeed a fantastic platform for online coaching and tuition services. Now, let's delve into your question about angular momentum.

We have two particles, each with mass mm and speed vv, traveling in opposite directions along parallel lines separated by a distance dd. To demonstrate that the vector angular momentum of this system remains the same regardless of the chosen point, let's denote the position vectors of the two particles as r1r1 and r2r2, respectively.

The angular momentum LL of a particle about a point OO is given by the cross product of its position vector rr and its linear momentum pp relative to that point:

LO=r×pLO=r×p

Now, let's calculate the angular momentum of each particle about an arbitrary point OO along their paths. For particle 1:

L1O=r1×p1L1O=r1×p1

And for particle 2:

L2O=r2×p2L2O=r2×p2

Since the particles are moving along parallel lines, their position vectors are parallel, and their magnitudes are equal, but their directions are opposite. Hence, r1=−r2r1=−r2.

Now, let's analyze the angular momentum of the system about point OO. The total angular momentum LtotalLtotal is the sum of the individual angular momenta:

Ltotal=L1O+L2OLtotal=L1O+L2O

Substituting the expressions for L1OL1O and L2OL2O, we get:

Ltotal=(r1×p1)+(r2×p2)Ltotal=(r1×p1)+(r2×p2)

Since r1=−r2r1=−r2, we can rewrite this as:

Ltotal=r1×(p1−p2)Ltotal=r1×(p1−p2)

Now, p1=mvp1=mv and p2=−mvp2=−mv (as the particles are moving in opposite directions), so:

p1−p2=mv−(−mv)=2mvp1−p2=mv−(−mv)=2mv

Substituting this back into the equation:

Ltotal=r1×(2mv)Ltotal=r1×(2mv)

Since r1r1 is the position vector of particle 1 relative to point OO, it doesn't change when we choose a different point. Therefore, LtotalLtotal remains the same regardless of the chosen point about which the angular momentum is calculated.

This demonstrates that the vector angular momentum of the two-particle system remains constant irrespective of the reference point chosen.

As a seasoned tutor registered on UrbanPro, I'd be delighted to assist you with this physics problem.

To solve this problem, we can use the concept of torque equilibrium. The sum of the torques acting on the bar must be zero for it to remain in equilibrium.

Let's denote the distance of the center of gravity of the bar from its left end as dd. Now, let's break down the forces acting on the bar:

1. The weight of the bar, WW, acts downward and can be considered to act at the center of gravity of the bar.
2. The tension forces in the strings exert forces on the bar at their points of contact.

Given that the bar is in equilibrium, we can write the condition for torque equilibrium about any point. Let's choose the left end of the bar as the pivot point.

The torque due to the weight of the bar about the left end is W×dW×d, and the torque due to the tension in the strings will depend on their distances from the pivot point.

Since the angles made by the strings with the vertical are given, we can use trigonometry to find the distances of the strings from the pivot point.

Let's denote the tensions in the strings as T1T1 and T2T2. Using the fact that the tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side, we can express the distances of the strings from the pivot point:

For the first string: tan⁡(36.9∘)=dxtan(36.9∘)=xd x=dtan⁡(36.9∘)x=tan(36.9∘)d

For the second string: tan⁡(53.2∘)=2−dytan(53.2∘)=y2−d y=2−dtan⁡(53.2∘)y=tan(53.2∘)2−d

Now, using the torque equilibrium condition:

T1×x=W×d=T2×yT1×x=W×d=T2×y

Substitute the expressions for xx and yy into the equation, and solve for dd. Once dd is found, it will give us the distance of the center of gravity of the bar from its left end.

I hope this helps! Let me know if you need further clarification or assistance with the calculations.

As an experienced tutor registered on UrbanPro, I can guide you through solving this physics problem step by step. First, let's break down the problem and identify the key concepts involved.

Given:

• Mass of the car (m) = 1800 kg
• Distance between front and back axles (L) = 1.8 m
• Center of gravity behind the front axle (d) = 1.05 m

To find:

• Force exerted by the level ground on each front wheel and each back wheel.

Now, the key concept here is the distribution of weight and the concept of torque. The weight of the car is acting downwards from its center of gravity, and it exerts a torque about the front and back axles. This torque causes the ground to exert a reaction force on the wheels.

First, let's find the torque exerted by the weight of the car about each axle. The torque (τ) is given by the formula:

τ=F⋅rτ=F⋅r

Where:

• F is the force (weight of the car)
• r is the perpendicular distance from the point of application of force to the pivot (axle in this case)

For the front axle: τfront=m⋅g⋅dτfront=m⋅g⋅d

For the back axle: τback=m⋅g⋅(L−d)τback=m⋅g⋅(L−d)

Where:

• m is the mass of the car (1800 kg)
• g is the acceleration due to gravity (9.8 m/s²)

Now, let's calculate these torques:

For the front axle: τfront=1800 kg×9.8 m/s2×1.05 mτfront=1800kg×9.8m/s2×1.05m

τfront=17640 N⋅mτfront=17640N⋅m

For the back axle: τback=1800 kg×9.8 m/s2×(1.8−1.05) mτback=1800kg×9.8m/s2×(1.8−1.05)m

τback=1800 kg×9.8 m/s2×0.75 mτback=1800kg×9.8m/s2×0.75m

τback=13230 N⋅mτback=13230N⋅m

Now, since the car is in equilibrium, the sum of torques about each axle must be zero.

Sum of torques=τfront−τbackSum of torques=τfront−τback

Sum of torques=17640 N⋅m−13230 N⋅mSum of torques=17640N⋅m−13230N⋅m

Sum of torques=4410 N⋅mSum of torques=4410N⋅m

This torque is balanced by the reaction forces at each axle.

Now, to find the force exerted by the level ground on each front wheel and each back wheel, we can use the fact that the torque exerted by a force about a pivot is equal to the force times the perpendicular distance from the pivot.

So, the force exerted by the ground on each wheel is:

For each front wheel: Ffront=Sum of torquesDistance between front wheelsFfront=Distance between front wheelsSum of torques

Ffront=4410 N⋅m1.8 mFfront=1.8m4410N⋅m

Ffront=2450 NFfront=2450N

For each back wheel: Fback=Sum of torquesDistance between back wheelsFback=Distance between back wheelsSum of torques

Fback=4410 N⋅m1.8 mFback=1.8m4410N⋅m

Fback=2450 NFback=2450N

Therefore, the force exerted by the level ground on each front wheel and each back wheel is 2450 N.

As a seasoned tutor registered on UrbanPro, I'm delighted to tackle this question. UrbanPro is indeed a fantastic platform for online coaching and tuition needs, offering a diverse range of subjects and experienced tutors.

Now, let's delve into the physics problem at hand. We're tasked with finding the moment of inertia of a sphere about a tangent to the sphere, given that the moment of inertia of the sphere about any of its diameters is 25MR252MR2, where MM is the mass of the sphere and RR is the radius of the sphere.

Firstly, let's recall a crucial property of moments of inertia. For a solid sphere rotating about an axis passing through its center, the moment of inertia II is given by 25MR252MR2, which is provided in the problem statement.

Now, when it comes to calculating the moment of inertia about a tangent to the sphere, we can utilize the parallel axis theorem. This theorem states that the moment of inertia about any axis parallel to an axis passing through the center of mass is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the perpendicular distance between the two axes.

In our case, the axis about which we know the moment of inertia is passing through the center of mass, and we want to find the moment of inertia about a tangent, which is parallel to this axis. Let's denote the moment of inertia about the tangent as I′I′.

So, using the parallel axis theorem:

I′=Icm+MD2I′=Icm+MD2

Where:

• Icm=25MR2Icm=52MR2 (moment of inertia about the center of mass)
• DD is the perpendicular distance between the two axes. In this case, D=RD=R, as the tangent is RR away from the center of the sphere.
• MM is the mass of the sphere.

Substituting the values:

I′=25MR2+MR2I′=52MR2+MR2

I′=75MR2I′=57MR2

So, the moment of inertia of the sphere about a tangent to the sphere is 75MR257MR2. This result should help you tackle similar problems in the future. If you have any further questions or need clarification, feel free to ask!

As an experienced tutor registered on UrbanPro, I'd like to help you with this physics problem. UrbanPro is an excellent platform for online coaching and tuition, where students can find knowledgeable tutors like myself to assist them with their academic needs.

Now, let's tackle your question about the moment of inertia of a disc. Given that the moment of inertia of a disc of mass MM and radius RR about its diameter is 14MR241MR2, we can find the moment of inertia about an axis normal to the disc passing through a point on its edge using the parallel axis theorem.

The parallel axis theorem states that the moment of inertia of a body about any axis parallel to an axis through its center of mass is equal to the sum of the moment of inertia about the parallel axis through the center of mass and the product of the mass of the body and the square of the distance between the two axes.

In this case, the distance between the axis passing through the center (where the moment of inertia is known) and the axis passing through a point on the edge is RR (since the radius of the disc is RR).

So, using the parallel axis theorem, we have:

I=Icenter+Md2I=Icenter+Md2

Where:

• II is the moment of inertia about the axis passing through a point on the edge,
• IcenterIcenter is the moment of inertia about the center axis (which is given as 14MR241MR2),
• MM is the mass of the disc,
• dd is the distance between the two axes, which is equal to RR.

Substituting the known values, we get:

I=14MR2+MR2I=41MR2+MR2

I=14MR2+44MR2I=41MR2+44MR2

I=54MR2I=45MR2

So, the moment of inertia about an axis normal to the disc passing through a point on its edge is 54MR245MR2.

If you need further clarification or assistance, feel free to ask!

As an experienced tutor registered on UrbanPro, let me guide you through this question. When it comes to analyzing rotational motion, we need to consider the principles of angular momentum and rotational inertia.

In this scenario, torques of equal magnitude are applied to both a hollow cylinder and a solid sphere, each with the same mass and radius. The cylinder rotates about its standard axis of symmetry, while the sphere rotates about an axis passing through its center.

To determine which object will acquire a greater angular speed after a given time, let's delve into their rotational properties:

1. Hollow Cylinder:

   • The rotational inertia (moment of inertia) of a hollow cylinder about its axis of symmetry is given by I=12MR2I=21MR2, where MM is the mass and RR is the radius.
   • Since the cylinder is free to rotate about its axis of symmetry, it can be treated as a solid cylinder for this analysis.
   • The torque applied will result in an angular acceleration, increasing its angular speed over time.

2. Solid Sphere:

   • The rotational inertia of a solid sphere about an axis passing through its center is given by I=25MR2I=52MR2.
   • Similar to the cylinder, the torque applied will lead to an angular acceleration, increasing its angular speed.

Comparing the rotational inertias, we can see that the solid sphere has a higher rotational inertia compared to the hollow cylinder, given the same mass and radius.

Now, according to the conservation of angular momentum, the angular momentum of each object will increase as the torque is applied. However, since the solid sphere has a higher rotational inertia, it will resist changes in its rotational motion more than the hollow cylinder. As a result, the solid sphere will acquire a lower angular speed compared to the hollow cylinder after the same amount of time under the influence of the same torque.

Therefore, the hollow cylinder will acquire a greater angular speed after a given time compared to the solid sphere when equal torques are applied to both.