guys i know the following 2 questions are very easy but i am not getting d same answer as d book has given..........pls help

1) stone mass 1kg in a vertical circle of radius 0.5m. find tension at lowest position, midway and highest position.

2)a 1kg block released from height 6m fricitionless track which ends in a quarter circle of radius 2m. what is the radial and tangential accn at pt q.............pt q is the end of the quarter circle

Let’s break down both of your questions step by step, ensuring we understand the concepts involved and how to arrive at the correct answers. We'll start with the first question about the stone in a vertical circle.

1. Tension in a Vertical Circle

When a mass is moving in a vertical circle, the tension in the string (or the force acting on the mass) varies depending on its position in the circle. We can analyze the forces acting on the stone at three key positions: the lowest point, the midpoint, and the highest point.

At the Lowest Position

At the lowest point of the circle, the forces acting on the stone are:

• The gravitational force acting downwards, which is equal to the weight of the stone: F_gravity = m * g
• The tension in the string acting upwards.

Using the formula for centripetal force, we know that the net force must equal the centripetal force required to keep the stone moving in a circle:

T - F_gravity = F_centripetal

Substituting the values:

• Mass (m) = 1 kg
• Acceleration due to gravity (g) = 9.81 m/s²
• Radius (r) = 0.5 m

First, calculate the gravitational force:

F_gravity = 1 kg * 9.81 m/s² = 9.81 N

Next, we need the centripetal force, which is given by:

F_centripetal = m * v² / r

To find the speed (v) at the lowest point, we can use energy conservation principles or assume it’s at maximum speed due to gravitational potential energy converting to kinetic energy. However, for simplicity, let's assume the stone is released from rest at the top of the circle (which we will calculate later). For now, we can say:

T = F_gravity + F_centripetal

At the Midpoint

At the midpoint, the gravitational force acts downwards, and the tension acts horizontally. The centripetal force is still required to keep the stone moving in a circular path. The tension here can be calculated as:

T = F_centripetal

At the Highest Position

At the highest point, the gravitational force acts downwards, and the tension also acts downwards. The equation becomes:

T + F_gravity = F_centripetal

Now, you can calculate the tension at each position once you have the speed at the lowest point. If you assume the stone starts from rest at the top, you can calculate the speed at the lowest point using energy conservation:

Potential Energy at Top = Kinetic Energy at Bottom

mgh = 0.5mv²

Solving for v gives you the speed at the lowest point, which you can then plug into the centripetal force equation to find the tensions.

2. Block on a Frictionless Track

For the second question, we need to analyze the motion of a block released from a height of 6 m that travels down a frictionless track and ends in a quarter circle of radius 2 m. We want to find the radial and tangential accelerations at point Q, which is at the end of the quarter circle.

Finding the Speed at Point Q

First, we calculate the speed of the block at point Q using energy conservation:

The initial potential energy when the block is at height 6 m is:

PE_initial = m * g * h = 1 kg * 9.81 m/s² * 6 m = 58.86 J

At point Q, the height is 2 m (the radius of the quarter circle), so the potential energy at point Q is:

PE_Q = m * g * h = 1 kg * 9.81 m/s² * 2 m = 19.62 J

The kinetic energy at point Q can be found by subtracting the potential energy at point Q from the initial potential energy:

KE_Q = PE_initial - PE_Q = 58.86 J - 19.62 J = 39.24 J

Now, using the kinetic energy formula:

KE = 0.5 * m * v²

We can solve for v:

39.24 J = 0.5 * 1 kg * v²

v² = 78.48

v = √78.48 ≈ 8.85 m/s

Calculating Radial and Tangential Acceleration

At point Q, the radial acceleration (a_radial) is given by:

a_radial = v² / r

Substituting the values:

a_radial = (8.85 m/s)² / 2 m ≈ 39.2 m/s²

The tangential acceleration (a_tangential) is zero in this case because there is no friction and the block is not gaining speed along the track at point Q; it is only changing direction.

In summary, at point Q:

• Radial acceleration: approximately 39.2 m/s²
• Tangential acceleration: 0 m/s²

By following these logical steps, you can arrive at the correct answers for both questions. If you have any further queries or need clarification on any part, feel free to ask!