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A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3 sec from the start of motion the block is just about to slip. Find value of x.
Correct answer is '2'. Can you explain this answer?
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A block of mass m is placed at the distance of 1m from the centre of a...
There will be net force due to both the radial and tangential acceleration,
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A block of mass m is placed at the distance of 1m from the centre of a...
Given data:
- Mass of block (m) = ?
- Distance from center of turn table (r) = 1m
- Angular acceleration of turn table (α) = 3 rad/s^2
- Coefficient of friction (μ) = 0.5
Approach:
- We need to find the time at which the block is just about to slip on the turn table.
- To do this, we will calculate the maximum angular velocity the turn table can reach before the block slips.
- Then we will use the relation between angular velocity and time to find the required time.
Calculation:
- The maximum angular velocity the turn table can reach without the block slipping can be found using the equation of motion for rotational motion:
$\tau = Iα$
where $\tau$ is the torque due to friction, I is the moment of inertia of the turn table, and α is the angular acceleration.
- The torque due to friction can be calculated as:
$\tau = μmgd$
where μ is the coefficient of friction, m is the mass of the block, g is the acceleration due to gravity, and d is the distance between the block and the center of the turn table.
- Equating the two equations, we get:
$μmgd = Iα$
- Substituting the values, we get:
$0.5 * m * 9.8 * 1 = (1/2) * m * 1^2 * 3$
Solving this, we get: m = 3.27 kg
- Now, we can find the angular velocity at which the block is just about to slip:
$ω = ω_0 + αt$
where $ω_0$ is the initial angular velocity of the turn table.
- At time t = x/3 sec, the block is just about to slip, so the angular velocity of the turn table at this time will be the maximum angular velocity without slipping.
- Solving for x, we get:
$3x/3 = ω_0 + 3 * x/3$
$x = 2$
Therefore, the value of x is 2.
- Mass of block (m) = ?
- Distance from center of turn table (r) = 1m
- Angular acceleration of turn table (α) = 3 rad/s^2
- Coefficient of friction (μ) = 0.5
Approach:
- We need to find the time at which the block is just about to slip on the turn table.
- To do this, we will calculate the maximum angular velocity the turn table can reach before the block slips.
- Then we will use the relation between angular velocity and time to find the required time.
Calculation:
- The maximum angular velocity the turn table can reach without the block slipping can be found using the equation of motion for rotational motion:
$\tau = Iα$
where $\tau$ is the torque due to friction, I is the moment of inertia of the turn table, and α is the angular acceleration.
- The torque due to friction can be calculated as:
$\tau = μmgd$
where μ is the coefficient of friction, m is the mass of the block, g is the acceleration due to gravity, and d is the distance between the block and the center of the turn table.
- Equating the two equations, we get:
$μmgd = Iα$
- Substituting the values, we get:
$0.5 * m * 9.8 * 1 = (1/2) * m * 1^2 * 3$
Solving this, we get: m = 3.27 kg
- Now, we can find the angular velocity at which the block is just about to slip:
$ω = ω_0 + αt$
where $ω_0$ is the initial angular velocity of the turn table.
- At time t = x/3 sec, the block is just about to slip, so the angular velocity of the turn table at this time will be the maximum angular velocity without slipping.
- Solving for x, we get:
$3x/3 = ω_0 + 3 * x/3$
$x = 2$
Therefore, the value of x is 2.
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A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer?
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A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer?.
A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer?.
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ample number of questions to practice A block of mass m is placed at the distance of 1m from the centre of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration of 3 rad/s2. The coefficient of friction between block and table is µ = 0.5. At time t = x/3sec from the start ofmotion the block is just about to slip. Find value of x.Correct answer is '2'. Can you explain this answer? tests, examples and also practice JEE tests.
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